Good quality notes on Physics - Magnetism. Very helpful for students, Study notes of Physics

Download Good quality notes on Physics - Magnetism. Very helpful for students and more Study notes Physics in PDF only on Docsity!   Digital Pvt. Ltd. [1] Concept of Field and Oersted Experiment Field Region around any physical quantity where another similar physical quantity experiences force or torque is called field. The branch of physics which deals with the magnetism due to electric current or moving charge (i.e. electric current is equivalent to the charges or electrons in motion) is called electromagnetism. Magnetic Field Region around a magnet or magnetic substance where another magnetic substance experiences force or torque is called Magnetic field. Unit : SI : weber/(meter)2 or tesla (T) C.G.S. : gauss (1 Tesla = 104 gauss) Dimensions : [MT–2A–1] 1T = 104 gauss Magnetism   Digital Pvt. Ltd. [2] Magnetism Source Field produced Magnet B Charge at rest E Current carrying wire B Charge in Uniform motion E,B An oscillating/accelerating charge produces: Electric Field Magnetic Field Electromagnetic Waves The nature of Force Illustration 1: A charge particle placed near to a current carrying wire then what is nature of force on rest charge? Solution: Electric force The charge particle will not experience magnetic force as it is stationary. Biot Savarts Law With the help of experimental results, Biot and Savart arrived at a mathematical expression that gives the magnetic field at some point in space in terms of the current that produces the field. That expression is based on the following experimental observations for the magnetic field dB at a point P associated with a length element d of a wire carrying a steady current I.   Digital Pvt. Ltd. [5] Magnetism Find Direction of Magnetic Field Easily!! Right Hand Thumb Rule Note: - For a straight current carrying wire M.F. lines are in form of concentric circles. Remember!! East, West, North and South are in Horizontal Plane Illustration 1: A wire has current in north direction find the direction of B due to it just below the wire Solution: The direction of magnetic field at point P will be towards west direction. I I ⦿ I I ⊗ This indicates a current carrying wire perpendicular to plane of paper in which current flows out of the paper This indicates a current carrying wire perpendicular to plane of paper in which current flows into the paper ⦿ N S W E Up ⦻ Down Up N E W S Down N i   Digital Pvt. Ltd. [6] Magnetism 1. Due to closed planar loops: • curl fingers of right hand, in the direction of current in the loop • Thumb will indicate the direction of magnetic field Magnetic Field due to Circular Coil Illustration 2: Which of the following is used to determine the direction of magnetic field in a current carrying conductor? Solution: The right-hand thumb rule determines the direction of magnetic field in a current carrying conductor. The rule states that when we align our right thumb in the direction of the current and curt our finders around it, the direction of our fingers is the direction of the magnetic field Illustration 3: The relation between the direction of current and the direction of magnetic field is Solution: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.   Digital Pvt. Ltd. [7] Magnetism Illustration 4: Find the direction of magnetic field at point R and S. Solution: Using Right Hand Palm Rule: - Magnetic Field due to moving Charge MF at a point due to a charge can be found in a way similar to that of a current element. MF due to current element ( )0 2 Id sin dB 4 r  =  Trick: ( )Id dq v→ MF due to moving charge ( )0 2 dq vsin dB 4 r  =  dq Id d dt   =     For moving charge ( )0 3 q v r B 4 r  =  0 2 qvsin B 4 r   =  ( ) d dq dt = ( )dq v= Direction: For Positive charge flow :- same as current element (Right Hand Rules) For Negative charge flow :- Opposite to current element. Illustration 1: A charge +3C moves vertically downward with speed 2 m/s. Find magnetic field at point P. Solution: 0 2 qv B 4 r  =  0µ (2)(4 ) B 4 4  =   = 0µ 2  Tesla   Digital Pvt. Ltd. [10] Magnetism So, net M.F. will be sum of all such elements 0 net Icos d B 4 a     =  here,  varies from 2− to 1 1 2 0 net I B cos d 4 a  −   =     M.F. at P due to straight current carrying wire 1 2 0 net I B cos d 4 a  −   =     ( ) 1 2 0 net I B sin | 4 a  −   =   ( )( )0 net 1 2 I B sin sin 4 a   =  − −  ( )0 net 1 2 I B sin sin 4 a   =  +   Direction : by right hand palm rule, In given figure, inwards Note: - In direction of current: 1 2Positive, Negative →  → In opposite direction of current: 1 2Negative, Positive →  →   Digital Pvt. Ltd. [11] Magnetism Illustration 1: Find magnetic field due to infinite length wire at point 'P'. Solution:  0 P I B sin90 sin90 4 d  =  +   0 P I B 2 d  =  Illustration 2: Find magnetic field due to given wire at point 'P'. Solution:  0 P I B sin sin90 4 d  =  +    0 P I B sin 1 4 d  =  +  Illustration 3: Find magnetic field due to semi-infinite length wire at point 'P'. Solution:  0 P I B sin0 sin90 4 d  =  +   0 P I B 4 d  =  P P P d 90° M I   Digital Pvt. Ltd. [12] Magnetism Magnetic Field due to Infinite Straight Wire Magnetic Field Due to Infiinte Wire 0 1 2 I B (sin sin ) 4 a  =  +   2 → Start Point of current 1 → End Point of current Here, 1 90 →  2 90 →  0 1 2 I B (sin sin ) 4 a  =  +   0I B (sin90 sin90 ) 4 a   =  +   0 0 0I I I B (1 1) (2) B 4 a 4 a 2 a      = + =  =    Illustration 1: Find magnetic field due to infinite length wire at point 'P'. Solution: 0 P I B 2 d  =  Illustration 2: Find B at P. Solution: 0 P I B 2 d  =  2 2d 3 4 5= + = 7 7 P 4 10 25 B 10 10 T 2 5 − −  = =   In vector form P P ˆ ˆB cos37i B sin37j− + 7 7ˆ ˆ8 10 i 6 10 j− −−  +  y P(3,4) X 25A (0,0) BP 37° d y P(3,4) X 25A (0,0)   Digital Pvt. Ltd. [15] Magnetism Illustration 1: If point ‘P’ lies out-side the line of wire then magnetic field at point ‘P’ will be : Solution: Here ( ) ( )1 1 2 290 90 =  −   =  −  ( ) ( )0 0 P 1 2 1 2 I I B sin 90 sin 90 (cos cos ) 4 d 4 d    = −  − −  =  −    Illustration 2: Find magnetic field due to finite length wire at point 'P'. Solution:  0 0 P P I I B sin sin0 ;B sin 4 d 4 d   =  +  =    Magnetic Field at the centre of Circular loop Direction: Magnetic field is Perpendicular to plane of circular loop   Digital Pvt. Ltd. [16] Magnetism Magnitude: M.F. at the center due to a current element 0 2 Id dB 4 R  =  Net M.F. at the center ( )0 0 0 2 2 I I B d 2 R 4 R 4 R   = =    If number of loops is not given, take single loop. For single loop : - 0 0 I B 2  =  For N loops : - 0 0 NI B 2R  = Illustration 1: Circular loop is made by a wire of length 7.5m. If current 5A is flowing in the loop, then find magnetic field at centre. Solution: 2 R 7.5 = 7.5 R 2 =  0 C I B 2R  = 7 24 10 5 8 2 7.5/2 −  =    2 740 10 T 15 −  Note:- If a single wire of constant length l is bent in form of N circular loops, Then, M.F. at the center due to current I flowing in the loops will be 0 0 0 0 NI NI NI1 2 N B 2R 2 R 2     = =  =  2B N  Illustration 2: A circular ring of 2 turns is made by a conducting wire. The magnetic field at the center is B. Now, this same wire is bent to form circular ring of 4 turns, find magnitude of new magnetic field at the center(current is same). R I R   Digital Pvt. Ltd. [17] Magnetism Solution: ( ) 2 2 1 1B N B 2 =  ( ) 2 2 2 2 2B N B 4 =  From above equation 2 1 2 2 B 2 B 4 = 2 1B 4B 4B= = Illustration 3: Two concentric circular coils of radius 5 cm and 10 cm are placed in same plane. If current 0.2 A is flowing in both coils in same direction, then find magnetic field at the centre. Solution: 0 1 1 I B 2r  = 0 2 2 I B 2r  = Net magnetic field at the centre (B0) = B1 + B2 = 0 1 2 I 1 1 2 r r   +    7 0 4 10 0.2 1 1 B 2 0.5 0.1 −    = +    7 0B 4.8 10 T−=  Illustration 5: Two concentric circular coils of radius 5 cm and 10 cm are placed in same plane. If current 0.2 A is flowing in both coils in opposite direction, then find magnetic field at the centre.   Digital Pvt. Ltd. [20] Magnetism Case II : Thickness ⎯⎯→ different, Material ⎯⎯→ same :– Magnetic field at the centre of arc abc and adc 0 1 0 2 abc adc I I B ,B 4r 4r   =  = Both arc are connected in parallel with source, so V = IR = constant and resistance of wire R = A  ( ) ( ) 1 1 I ..... 1 R ..... 2 R A   From (1) & (2) thick thinI A I I   Net magnetic field at centre 0 0 net 1 2(B ) (I –I ) 4r  =  (where I1 > I2) Illustration 1: For the arrangement made up of two identical coils as shown in figure, determine the magnetic field at the centre O. Solution: The two coils are perpendicular to each other. Coil 1 produce field along X axis and coil 2 produce field along y axis. thus, the resultant field will be 2 2 1 2B B B= + Making an angle  from x-axis 2 1 B tan B  = For identical coils, 0 1 2 NI B B ;B 2 2a   = =     and hence 45 =  Illustration 2: The magnetic induction at the centre O is? Solution: 0 0I I3 1 B 4 2a 4 2b      =  +         0 03 I I B 8a 8b    = +     I1 Thick O a I  Θ I2 d b Thin r c a B1 O o X B Y B2 Coil 2 Coil 1   Digital Pvt. Ltd. [21] Magnetism Illustration 3: A wire is bent in the form of a circular arc with a straight portion AB. Magnetic induction at O when current I flowing in the wire, is? Solution: 0 AB I B [2sin ] 4 (OC)  =   But OC rcos=  or 0 AB I B tan 2 r  =   Magnetic field due to circular portion, 0 0 AB I I2 2 B ( ) 2r 2 2 r   −   = =  −      Total magnetic field = 0 0 0I I I tan ( ) [tan ] 2 r 2 r 2 r     +  −  =  +  −     Illustration 4: A long wire bent as shown in the figure carries current I. If the radius of the semi-circular portion is "a" then find the magnetic induction at the centre C. Solution: Due to semi-circular part ( )0 1 I ˆB i 4a  = − Due to parallel parts of currents 0 0 0 2 net C 1 2 I I Iˆ ˆˆB 2 ( k),B B B B ( i) ( k) 4 a 4a 2 a    =  − = = + = − + −   Magnitude of resultant field 2 2 20 net 1 2 I B B B 4 4 a  = + =  +  Illustration 5: A current i ampere flows in a circular arc of wire whose radius is 2R, which subtend an angle3 / 2 radian at its centre. The magnetic induction B at the centre is Solution: 0 0 0 2 i 3 i(2 )i 2 B 4 R' 4 2R 16R    −     −   = = =   O 2R /2 i O I A B C r r O O   Digital Pvt. Ltd. [22] Magnetism Illustration 6: Figure shows a circular loop with radius 'r'. The resistance of arc ABC is 2 and that of ADC is 4 . Magnetic field at the centre of the loop is: - Solution: Magnetic field at the centre due to arc ABC and ADC :- BABC = 0 1I 4r   , BADC = 0 2I 4r  Both arc are connected in parallel with source, so V= IR = constant 2 1 1 2 I(R ) 4 2I I = = I = (R +R ) 6 3       1 2 1 2 IR I I = = R +R 3 Net magnetic field at the centre :- (B0)net = 0 4r  (1 – 2)= 0 4r  2I I – 3 3       = 0I 12r   Special Case: When current divides in any symmetrical planar loop made with uniform wire, then magnetic field at the centre due to this loop is ZERO. Magnetic Field at the axis of Circular loop Magnetic Field on the Axis of Current Carrying circular Loop Direction Component of magnetic field perpendicular to axis of ring will be zero. Because the components of magnetic field produced due to opposite current elements on the ring cancel each other. Net magnetic field due to ring at any point on the axis will be along the axis. B   A O r D Θ 1 2 C B   A O r D Θ C   Digital Pvt. Ltd. [25] Magnetism Graph : B vs x At P and Q, graph changes its curvature P and Q are known as • Points of Inflection or • Points of Zero curvature or • Points of Curvature change. Helmholtz's Coil Used to produce Uniform magnetic field (constant magnitude and direction) in a short range (in mm). 0NI B 0.72 R   Conditions 1. Both coils must be coaxial 2. Same current in same sense 3. Equal number of turns and equal radius for both coils 4. Separation must be equal to R Illustration 1: A coil having N turns carry a current I as shown in the figure. The magnetic field intensity at point P is I X=4 P R R   Digital Pvt. Ltd. [26] Magnetism Solution: Magnetic field on the axis of circular coil or loop = 2 0 2 2 3/2 NIR 2(R x )  + Illustration 2: Radius of a current carrying coil is ‘R’. The ratio of magnetic field at a axial point which is 3R distance away from the centre of the coil to the magnetic field at the centre of the coil: - Solution: 0 x 3/22 2 B B x 1 R =   +    x = 3R  0 0 x 3/2 3/2 B B B (1 9) (10) = = +  3/2 x 0 B 1 B 10   =     3/2 x 0 1 B B 10   =     Illustration 3: Radius of current carrying coil is 'R'. If fractional decreases in field value with respect to centre of the coil for a nearby axial point is 3% then find axial position of that point. Solution: Decrease in field value with respect to centre X 0 XB B B = − = B0 – B0 2 2 3 x 1 2 R   −    = B0 2 2 3 x . 2 R       Fractional decrease in field value (Where x <<< R) 2 x 2 0 B 3 x B 2 R  = (limit  5%)  2 2 3 3 x 100 2 R =  R x 50 =  Illustration 4: Magnetic fields at two points on the axis of a circular coil at a distance of 0.04m and 0.2m from the centre are in the ratio 8 : 1. The radius of the coil is Solution: 2 0 2 2 3/2 2 2 3/2 2 NiR 1 B B 4 (R x ) (r x )   =     + + 2/32 2 3/2 2 2 2 2 3/2 2 1 (R x )8 8 R 0.04 1 1(R x ) R 0.0016 + +   =  =  + +  2 2 4 R 0.04 1 R 0.0016 +  = + On solving R = 0.10 m   Digital Pvt. Ltd. [27] Magnetism Ampere’s Circuital Law and its basic applications Ampere Circuital Law It gives another method to calculate the magnetic field due to given current distribution. Line integral of magnetic field along any closed path is equal to µ0 times of enclosed current. 0 encB d I =  Note: 1. It is applicable only for steady/constant current 2. It can be applied for any distribution of current but, 3. It is applied for symmetric distribution only for calculation purpose 4. Any shape of loop is acceptable but, 5. B and d should be parallel or perpendicular to each other. Magnetic Field due to Long Thin wire 0 encB d I =  0B(2 r) (I) =  0(I) B 2 r  =  0 I B 2 r  =  Magnetic Field due to Long Thick wire At a point outside wire 0 encB d I =  0B(2 r) (I) =  0(I) B 2 r  =  0 out I B 2 r  =  , 0 sur I B 2 R  =  At a point inside wire 0 encB d I =  2 0 2 I B(2 r) r R    =       0 2 (Ir) B 2 R  =  I3 I7 I1 I2 I5 I4 I6 I ∞ ∞ I r r   Digital Pvt. Ltd. [30] Magnetism Illustration 4: A long straight wire (radius = 3.0 mm) carries a constant current distributed uniformly over a cross section perpendicular to the axis of the wire. If the current density is 50 A/m2. The magnitudes of the magnetic field at (a) 1.5 mm from the axis of the wire and (b) 4.0 mm from the axis of the wire is :- Solution: (a) Magnetic field inside the conductor: - Bin = 0 2 Ir 2 R   = 0 2 r I 2 R        = 0r 2  (J), (Where J= 2 I R current density) = –7 –34 10 1.5 10 50 2      = 1.5 × 10–8 T (b) Magnetic field outside the conductor :- Bout = 0I 2 r   [ J = 2 I R  I = J( 2R )] = 2 0 J( R ) 2 r    = 2 0 JR 2r  = –7 –3 2 –3 4 10 50 (3 10 ) 2 4 10        = 9 4  × 10–8 T Magnetic field due to Solenoid Solenoid ➢ It consists of a conducting wire which is tightly wound over a cylindrical frame in the form of helix. All the adjacent turns are electrically insulated to each other. The magnetic field at a point on the axis of a solenoid can be obtained by superposition of field due to large number of identical circular turns having their centres on the axis of solenoid. ➢ M.F. is produced along Axis of solenoid. Outside a long solenoid, M.F. is almost Zero. ➢ Direction of magnetic field ( )B inside a solenoid can be obtained by using Right Hand Thumb Rule. (a) (b) Magnetic Field Inside A Long Solenoid   Digital Pvt. Ltd. [31] Magnetism For the loop PQRS, we have PQ QR RS SP B d B d B d B d B d =  +  +  +      Here, ( ) PQ RS B.d B.d 0 B d = = ⊥  and SP B.d 0= ( B outside the solenoid is negligible) and QR QR B d B d Ba = =  Now, B d Ba = Current enclosed for loop PQRS, ( )i n a i =   where n = number of turns per unit length and i = current through wire of solenoid From Ampere's Circuital Law for the loop PQRS, we have 0B d i =   ( )0B a n a i  =    0B ni =  Thus, magnetic field inside a long solenoid. General Parameters of Solenoid N : Number of Turns : Length of Solenoid n : Number of Turns per unit length Hence N n = R : Radius of Solenoid coil r : Relative permeability of core of solenoid Magnetic Field due to Long Solenoid For a long solenoid, carrying current I M.F. in the central region 0 0B nI=  0 0 N B I   =      M.F. outside the solenoid B = 0 M.F. at the edges/end points 0n I B 2  = In presence of Core, 0 0 r →   Where 𝜇𝑟 is the relative permeability of core material. Magnetic Field due to Finite Length Solenoid ( )0 P 1 2 n I B cos cos 2  =  +  B   Digital Pvt. Ltd. [32] Magnetism Note: Magnetic field outside a finite length solenoid is not zero. For long solenoid, in the middle region o 1 2, 0  → ( )o o 0 cos0 cos0 B nI 2 +  =  0B nI =  For long solenoid, in the end region o o 1 2, 0 , 90  → ( )o o 0 cos0 cos90 B nI 2 +  =  0nI B 2   = Key Note Magnetic Field produced by a Solenoid resembles the Magnetic field of a Bar Magnet So, a solenoid acts as a Magnetic Dipole and is used as Electromagnet in many devices. Solenoid Lock Mechanism used in many Automatic Door Locks and other electro-mechanical equipment Illustration 1: The length of solenoid is 0.1m and its diameter is very small. A wire is wound over it in two layers. The number of turns in inner layer is 50 and that of outer layer is 40. The strength of current flowing in two layers in opposite direction is 3A. Then find magnetic induction at the middle of the solenoid. Solution: Direction of magnetic field due to both layers is opposite, as direction of current is opposite, so ( )1 2 net 1 2 0 1 1 0 2 2 0 0 1 2 N N B B B n I n I I I I I I= − =  −  =  −  = = = ( ) ( ) 7 50 1 2 I 4 10 3 N N 50 40 12 10 T 0.1 − −   − = − =    Digital Pvt. Ltd. [35] Magnetism Solution: We are given that, mean radius of the toroid, i.e., r = 25cm 26cm 2 + = 25.5 cm = 0.255 m total number of turns, N = 3500 current through the toroid, I = 11 A Clearly number of turns per unit length of the toroid, n = N 3500 2 r 2 0.255 =   (a) Magnetic field outside the toroid is zero as it non-zero inside its core. (b) Magnetic field inside the core of the toroid, i.e., 7 2 0 3500 B nI (4 10 ) 11 T 3.0 10 T 2 0.255 − −   =  =   =       Note : Inside the toroid, 1 B r  . The above value of B corresponds to mean radius 25.5 cm. (c) In the empty space surrounded by the toroid, B = 0 Key Note • Magnetic field of toroid is non-uniform because at each point direction of magnetic field is not same. • It is not a Magnetic Dipole • Magnetic field produced by a toroid is directed along its circular axis and constant in magnitude over its entire cross section. • Magnetic field outside the volume of toroid is always zero. • Magnetic field at the centre of toroid is always zero. Magnetic Force on Moving Charge Motion of Charge in Magnetic Field Consider a charge q moving with velocity v in uniform magnetic field B such that v makes an angle  with B , then magnetic force experienced by the charge is given by mF q(v B)=  ( )F q V B=  F V⊥ F B⊥ B R1 R2   Digital Pvt. Ltd. [36] Magnetism Direction of force on a positive charge find out with help of cross product and Right-Hand Rule. ( )F q V B=  F = qvB sin   = 0°  can vary from 0° to 180°  = 180° F = 0 ( )0 & 180 =  =  Illustration 1: A test charge 1·6 × 10–19 Cb is moving with velocity ˆ ˆv =(2i + 3j) m/sec is a magnetic field ˆ ˆB = (2i + 3j) Wb/m2. The magnetic force on the test charge: - Solution: Magnetic force on moving charge :- Here :-  v || B  0 =  Fm = qvBsin   Fm = 0 Illustration 2: An electron is moving along +x direction. To get it moving along an anticlockwise circular path in x– y plane, magnetic field applied along :- Solution: Sense of path gives an idea of direction of magnetic force (i.e. centripetal force), along +y–direction. Now direction of external magnetic field can be find out by right hand palm rule. Hence magnetic field direction along +z–direction. B 𝐅Ԧ 𝐁ሬሬԦ 𝐯ሬԦ B B y-axis Fm x-axis v e– (ACW in x-y plane) B   Digital Pvt. Ltd. [37] Magnetism Illustration 3: A proton moving along z-axis with constant velocity. If a magnetic field is applied along y-axis then direction of magnetic force on proton Solution: mF → = ± q(v B) mF  + ( k̂ × ĵ ) = – î ,  along – x direction Illustration 4: Find direction of & plane of motion. Solution: ( )F = q V B ( )ˆˆ ˆj i k = − ( )ˆF k→ −  y z plane Illustration 5: From the path followed, identify nature of charge on the particle. Solution: ( )F = q V B (1) Positive charge = According to Right-Hand Rule positive charge will be move in circular upward direction. (2) Neutral  q = 0 F = 0 No magnetic force No deviation (3) Negative charge = According to Right-Hand Rule negative charge will be move in circular downward direction. B 1 2 3   Digital Pvt. Ltd. [40] Magnetism Solution: 1. mv R qB = Particle Mass Charge Proton mp +e Deutron 2 mp +e Alpha 4 mp +2e m R q  31 2 1 2 3 1 2 3 mm m R :R :R : : q q q = =1 : 2 : 2 2. p R qB = 1 R q  (P → same) 1 2 3 1 2 3 1 1 1 R :R :R : : q q q = = 1 : 1 : 1 2 = 2 : 2 : 1 3. 2mK R qB = (K → same) m R q  31 2 1 2 3 1 2 3 mm m R :R :R : : q q q = = 1 : 2 : 1 4. acc2mV1 R B q = m R q  Vacc.→ same 31 2 1 2 3 1 2 3 mm m R :R :R : : q q q = = 1 : 2 : 2 = 1 : 2 : 2 Illustration 2: A Deuteron of kinetic energy 100 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field B . The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is   Digital Pvt. Ltd. [41] Magnetism Solution: 22mK q r K qB m =   2 2 p p d d d p K q m 1 2 2 K q m 1 1 1      =  =  =       pk 2 100 200 keV. =  = Motion of Charge Particle in Limited Magnetic Field For general angle of entrance  1. Deviation angle 2 2. Time spent in M.F. 2 T' T 2  =   where 2 m T qB  = Or 2 T' T 360  =   3. Displacement of particle s 2rsin =  A uniform magnetic field of width 'd' shown in figure. A charge (q,m) is projected from field free region to field region in such a way that its velocity (v) is perpendicular to the magnetic field (B). The charge deflected from its initial direction of motion and trace a part of circle according to following conditions: - (a) Charge does not strike to the opposite plate and completes the semi-circle  r < d (b) Charge does not strike to the opposite plate and just completes the semi-circle  r = d (c) Charge strikes to the opposite plate and does not completes the semi-circle  r > d Illustration 1: A charge particle (q, m) enters in transverse B as shown in fig. Then find- (1) Deviation angle (2) Time for which particle remains in M.F. (3) Displacement of particle when particle emerges out. B R C × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × 45° × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × V B   Digital Pvt. Ltd. [42] Magnetism Solution: (1) Deviation angle = 45°+45° = 90° (2) Time for which particle remains in M.F. : - We know that t =   2 t 2 T   =   2 t 2 m2 qB   =    m t 2qB  = (3) Displacement of particle when particle emerges out : -  Rcos45 Rcos45 +   2R mv R qB   =     mv 2 qB Illustration 2: A particle of mass m and charge q is projected into a region having a perpendicular uniform magnetic field B of width d. Find the angle of deviation q of the particle as it comes out of the magnetic field. Solution: The radius of the circular orbit is r = mv qB The deviation  may be obtained from the fig. as 1d dBq dBq sin or sin r mv mv −    = =  =     Illustration 3: A proton accelerated by a potential difference 500 kV moves though a transverse magnetic field of 0.51T as shown in figure. The angle  through which the proton deviates from the initial direction of its motion, is R R   Digital Pvt. Ltd. [45] Magnetism Pitch of helix (p) The linear distance travelled by the charge particle in one revolution or in one time period along external magnetic field direction is called 'pitch of helix'. Displacement in T time p = (v cos)T , where T = 2 m qB   2 mvcos p qB   = Illustration 1: A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be Solution: When particle enters at an angle other than 0º, 90º or 180º, path followed is helix. Illustration 2: A charge (q, m) projected with velocity ˆ ˆv 3i 4j= + m/s in external magnetic field ˆB 6iT= . Find (i) Radius of circular path (ii) Pitch (iii) Time period Solution: 4 sin 5  = Radius of circular path. mvsin R qB  = 4 m 4 5R q 6   =  Time period = 2 m 6q  Pitch = p = (v cos)T p = 3 × 2 m 6q  Illustration 3: A charged particle enters a uniform magnetic field with velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be   Digital Pvt. Ltd. [46] Magnetism Solution: 2 m 2 m p (vcos45 ) (vsin45 ) Bq Bq   =  =   mvsin45 p Bq 2  =  = radius of helix Illustration 4: A proton of mass 1.67 × 10–27 kg and charge 1.6 × 10–19 C is projected with a speed of 2 × 106 m/s at an angle of 60° to the X-axis. if a uniform magnetic field of 0.104 T is applied along Y-axis, the path of proton is Solution: Path of the proton will be a helix of radius, mvsin r qB  = (where  = angle between B and v )  27 6 19 1.67 10 2 10 sin30 r 1.6 10 0.104 − −      =   = 0.1 m Time period, 27 19 2 m 2 1.67 10 T qB 1.6 10 0.104 − −    = =   = 2 × 10–7 s Lorentz Force When a charge is moving in a region, where both electric field E and magnetic field B exist, then electric and magnetic forces are acting on it. The resultant of these forces called electromagnetic or Lorentz force on charge. (zero gravity) Net Electromagnetic force is known as Lorentz force. net m eF F F= + ( ) ( )netF q v B qE=  + • If Fnet = 0 Then v = constant Note : If E 0 , B 0  and netF 0= ( ) ( )q v B qE 0 + = ( )E v B= −  for both positive and negative charges Illustration 1: A beam of protons is deflected sideways. Could this deflection be caused by (i) a magnetic field (ii) an electric field? If either possible, what would be the difference?   Digital Pvt. Ltd. [47] Magnetism Solution: Yes, the moving charged particle (e.g. proton, a-particles etc.) may be deflected sideway either by an electric or by a magnetic field. (i) The force exerted by a magnetic field on the moving charged particle is always perpendicular to direction of motion, so that no work is done on the particle by this magnetic force. That is the magnetic field simply deflects the particle and does not increases its kinetic energy. (ii) The force exerted by electric field on the charged particle at rest or in motion is always along the direction of field and the kinetic energy of the particle changes. Illustration 2: A particle moving with velocity of v as shown in figure then the path of trajectory and also find the effect on the speed. Solution: Path will be straight line with increasing speed. Illustration 3: mutually ⊥ Find v such that particle moves constant velocity. Solution: qE = qvB E v B = Illustration 4: A particle of charge –1.6 × 10–18 coulomb moving with velocity 10 ms–1 along the x-axis enters a region where a magnetic field of induction B is along the y-axis, and an electric field of magnitude 104 V/m is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is Solution: Since particle is moving undeflected. So 4 3 210 qE qvB B E / v 10 Wb / m 10 =  = = = Illustration 5: An electron (mass = 9.1 × 10–31 kg; charge = 1.6 ×10–19 C) experiences no deflection if subjected to an electric field of 3.2 × 105 V/m, and a magnetic field of 2.0 × 10–3 Wb/m2. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius   Digital Pvt. Ltd. [50] Magnetism Note : • Time period does not depend on velocity or Radius!! • It depends on ratio of charge to mass of the particle Particle Charge Mass q/m electron e me = 9.1 × 10−31 kg 0.17 × 1012 C/kg proton e mp = 1.67 × 10−27 kg 0.95 × 108 C/kg neutron 0 mp  mp 0 deuteron e 2mp 0.47 × 108 C/kg  -particle 2e 4mp 0.47 × 108 C/kg • Charge to mass ratio is known as specific charge. Illustration 8: Four charge particles proton, electron, neutron and a-particle projected with same velocity in same uniform transverse magnetic field identify according to the given diagram. mv R qB = Solution: mv R qB = …..(1) m R q  31 2 4 1 2 3 4 1 2 3 4 mm m m R :R :R :R :: : : : q q q q Deflection of particle from initial direction ( )  Curvature (C) 1 Radius of Curvature (R)  …..(2) We identify in following graph with help of Relation (1) and (2) 1 → graph for  – particle 2 → graph for proton 3 → graph for neutron 4 → graph for electron 1 2 3 4 q 1 q 2 q 3 q 4   Digital Pvt. Ltd. [51] Magnetism Illustration 9: He2+ and Cl–, both are projected with same momentum in a transverse magnetic field. Which one will have more curvature? Solution: m R q  Deflection of particle from initial direction ( )  Curvature (C) 1 Radius of Curvature (R)  – 2Cl He m m q q +             So, He2+ has more curvature. Cyclotron The cyclotron is a device which is used to accelerate the charge particles or ions to high energies. It was invented by E.O. Lawrence and M.S. Livingston in 1934. Principle When a charged particle or ion is made to move again and again in a high frequency electric field and strong magnetic field, it gets accelerated and acquires sufficiently large amount of energy. Construction It consists of two hollow D–shaped metallic chambers D1 and D2 called dees. These dees are separated by a small gap, where a source (P) of charged particles or ions is placed. Dees are connected to high frequency oscillator, which provides high frequency electric field across the gap of the dees. This arrangement is placed between the two poles of a strong electromagnet. The magnetic field due to this electromagnet is perpendicular to the plane of the dees. Working A charge particle of mass 'm' and charge 'q' moves at right angle to the magnetic field 'B' inside the dees. (i) Radius of circular path traversed by the charge: – m r qB  = (ii) Time taken by the charge to complete the semi-circle: – m t qB  = This shows that time taken by the charge to complete any semi-circle is same. (iii) The time period of alternating electric field is 'T'. The polarities of dee will change after time 'T/2'. The charge will be accelerated if: – T m t 2 qB  = = (Resonance condition) 2 m T qB   = (iv) The frequency of alternating electric field: - 1 qB v T 2 m =   =  B D2 D1 He2 + C l– q 1 q 2   Digital Pvt. Ltd. [52] Magnetism (v) Maximum kinetic energy gained by the charge: - 2 2 2 K max. max. q B (E ) r 2m   =     where rmax. is radius of the longest semi-circle described by the ion. (vi) Total energy acquired by the charge during ‘N’ complete revolution: TE 2NqV= , where V is potential difference to which the charge is subjected every time, it enters a dee. This energy must be equal to the maximum kinetic energy of charge: 2 2 2 2 2 max maxB q r B qr 2NqV N 2m 4Vm =  = Limitations of Cyclotron (i) Cyclotron is not used to accelerate uncharged particles like neutron. (ii) Cyclotron is not used to accelerates electrons because they have very small mass. It is used to speed up POSITIVE CHARGED PARTICLES to acquire high kinetic energy for nuclear reactions. It is used to speed up Positive Charged Particles to acquire high kinetic energy for nuclear reactions. B B D2 D1 B D2 D1 Electric Field Oscillator   Digital Pvt. Ltd. [55] Magnetism Leff is the displacement vector from starting point of current to end point of current. Example For straight wire, Arbitrary shape current carrying conductor For loop, | Leff | = L | Leff | = 0  Current element in a magnetic field does not experience any force if the current in it is parallel or anti–parallel with the field  = 0° or 180° dFm = 0 (min.)  Current element in a magnetic field experiences maximum force if the current in it is perpendicular with the field  = 90° dFm = BId (max.)  Magnetic force on current element is always perpendicular to the current element vector and magnetic field vector. mdF → ⊥ Id and mdF → ⊥ B (always) Direction of magnetic force Magnetic force on the wire we can find by Right Hand Palm Rule Fingers → In direction of external Magnetic field Thumb → In direction of L ⊥ to Palm → Direction of magnetic force on wire Note: For straight wire, direction of L is in the direction of current flow. Illustration 1: Find force on given wire in uniform magnetic field 𝐅Ԧ 𝐁ሬԦ 𝐋Ԧ   Digital Pvt. Ltd. [56] Magnetism Solution: ( )m effF I L B =  = I(Leff)B = I(2R)B (Upward) Illustration 2: A current of 10 amperes is flowing in a wire of length 1.5 m. A force of 15 N acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of the current is Solution: F 15 1 F Bilsin sin Bil 2 10 1.5 2 =    = = =     = 30° Illustration 3: A wire bent as shown in fig carries a current i and is placed in a uniform field of magnetic induction B that emerges from the plane of the figure. Calculate the force acting on the wire. Solution: The total force on the whole wire is Fm = | L |B = (R + 2R + R)B = 4RB Illustration 4: Find force on given wire. Solution: Leff = 10 m Fm = Leff B = 20 × 10–3 × 10 × 10 = 2N Illustration 5: A current 'I' is flowing through wire PQR. This wire is bent in form of an angle and placed in uniform magnetic field 'B' according to figure. If PQ = and PQR = 60°. The ratio of magnetic forces on PQ to QR respectively: -   Digital Pvt. Ltd. [57] Magnetism Solution: FPQ = I B sin = I B sin90° = I B ...(1) FQR = I 'B sin30° =  cos60 B sin30° = I(2 )B × 1 2 = I B ...(2) from (1) & (2) PQ QR F F = I B I B = 1 : 1 Current Carrying Loop in Uniform Magnetic Field Force : Net force on a closed loop(any shape) in uniform magnetic field is always ZERO. Tension Developed in the wire : T = RBI If length of wire is given 2 R R 2 =   =  Note : Due to Tension in wire, loop of any shape becomes circular in shape.   Digital Pvt. Ltd. [60] Magnetism Illustration 2: Two long and parallel wires are at a distance of 0.1 m and a current of 5 A is flowing in each of these wires. The force per unit length due to these wires will be Solution: 7 50 1 22 i i 10 2 5 5 F 5 10 N / m 4 a 0.1 − −     = = =   Illustration 3: A current of 5A flows through two long parallel wires. The magnetic force on each wire is 4× 10–3 N/m. If their currents makes half and separation between them makes double then magnetic force per unit length of each wire becomes :- Solution: Magnetic force per unit length fm = 0 1 2I I 2 d   .....(1) f'm = ( ) 1 2 0 I I 2 2 2 2d            = 0 1 2I I 1 2 d 8    .....(2) from (1) & (2) f'm = mf 8 = 3 34 10 0.5 10 N / m 8 − − =  Illustration 4: Three long straight wires, carrying currents are arranged according to figure. Magnetic force on 20cm part of the wire B is :- Solution: Resultant force on ' ' length of wire B Fnet = (FC – FA) , (FC > FA) = 0 0(8)(6) (3)(6) – 2 (4) 2 (2)         × 20 × 10–2 = 1.2 × 10–7 N, towards right Illustration 5: Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force experienced by 10 cm length of wire Q is   Digital Pvt. Ltd. [61] Magnetism Solution: Force on wire Q due to wire P is 7 5 P 2 30 10 F 10 0.1 6 10 N 0.1 − −  =   =  (Towards left) Force on wire Q due to wire R is 7 5 R 2 20 10 F 10 0.1 20 10 N 0.02 − −  =   =  (Towards right) Hence 5 4 net R PF F F 14 10 N 1.4 10 N− −= − =  =  (Towards right) Illustration 6: What is the net force on the square coil Solution: Force on side BC and AD are equal but opposite so their net will be zero. But 7 2 6 AB 2 2 2 1 F 10 15 10 3 10 N 2 10 − − − −   =    =   and ( ) 7 2 6 CD 2 2 2 1 F 10 15 10 0.5 10 N 12 10 − − − −   =    =    6 net AB CDF F F 2.5 10 N−= − =  725 10 N−=  , towards the wire. Equilibrium of free wire Case I : Upper wire is free : Consider a long horizontal wire which is rigidly fixed another wire is placed directly above and parallel to fixed wire. Magnetic force per unit length of free wire fm = 0 1 2I I 2 h   , and it is repulsive in nature because currents are unlike. Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit length At balanced condition fm = W'. Weight per unit length of free wire = 0 1 2I I m g 2 h  =  (stable equilibrium condition) 2A 2cm 15cm 10cm A FAB FCD B C D 1A   Digital Pvt. Ltd. [62] Magnetism If free wire is slightly displaced and released then it will executes S.H.M. in vertical plane. The time period of motion is h T 2 g =  Case II : Lower wire is free : Consider a long horizontal wire which is rigidly fixed. Another wire is placed directly below and parallel to the fixed wire. Magnetic force per unit length of free wire is fm = 0 1 2I I 2 d   , and it is attractive in nature because currents are like. Free wire may remains suspended if the magnetic force per unit length is equal to weight of its unit length At balanced condition fm = W' Weight per unit length of free wire 0 1 2I I m g 2 d  =  (unstable equilibrium condition) Illustration 7: A long horizontal wire is rigidly fixed and carries 100A current. Another wire of linear mass density 2 × 10–3 kg/m placed below and parallel to the fixed wire. If the free wire kept 2cm below and hangs in air, then current in free wire is:- Solution: At balanced condition of free wire :- fm = g 0 1 2I I 2 d   = g 2 = 0 1 2 gd I   ( = 2 × 10–3 kg/m) = 3 2 7 2 3.14 2 10 9.8 2 10 4 10 100 − − −          = 19.6 A (in free wire) Bar Magnet and Magnetic Dipole Bar Magnet A bar magnet is a rectangular piece of the object. It is made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, having permanent magnetic properties. Two poles are present in a Bar magnet, North Pole and South Pole. N S   Digital Pvt. Ltd. [65] Magnetism Magnetic moment of each part M' = m( /2) M M' 2 = Sp. Note :- A bar magnet of magnetic moment ‘M’ cuts into ‘n’ equal parts then magnetic moment of each part becomes : M M' n = Illustration 2: A straight steel wire of length ' ' has a magnetic moment ‘M’. It is bent into various shapes shown in figure. Find out new magnetic moment of each shape. (a) (b) (c) (d) Solution: If a flexible magnet of magnetic moment ‘M’ is bent into any shape then magnetic moment of new shape is always decreases as effective length is always decreases (but pole strength remains same) (a) M' = m ( )2 M' = M 2 (b) M' = m( /2) M' = M 2   Digital Pvt. Ltd. [66] Magnetism (c) M' = m(2R) where = R M' = 2m( /) M' = 2M      (d) Effective length of system is zero so magnetic moment is also zero. Illustration 3: Two/three/four identical bar magnets of magnetic moment 'M' are combined according to figure. Find net magnetic moment of the system. (a) (b) Solution: (a)  = 120°  Mnet = 2 2m M M 2MM cos120+ +  M (b) According to polygon rule of vector addition if all vector can be represented by sides of polygon in sequence the resultant (sum) of all vectors is zero so Mnet = 0   Digital Pvt. Ltd. [67] Magnetism Magnetic Moment of Current Carrying Coil (Loop) Current carrying coil (or loop) behaves like magnetic dipole. The face of coil in which current appears to flow anticlockwise acts as north pole while face of coil in which current appears to flow clock wise acts as south pole. A current carrying coil acts as a Magnetic Dipole ACW → North Pole CW → South Pole • A loop of geometrical area 'A', carries a current '' then magnetic moment of coil M = A • A coil of turns 'N', geometrical area 'A', carries a current 'I' then magnetic moment M = N  A Magnetic moment of current carrying coil is an axial vector M NIA= where A is a area vector perpendicular to the plane of the coil and along its axis. SI unit : A-m2 or J/T Direction of M find out by right hand thumb rule • Curling fingers  In the direction of current • Thumb  Gives the direction of M For a current carrying coil, its magnetic moment and magnetic field vectors both are parallel axial vectors. Illustration 4: A wire of length ' ' carries a current bent into various regular polygons shown below. Find magnetic moment in terms of length of the wire. (a) Equilateral triangle (b) Square (c) Regular hexagon (d) Circular loop Solution: (a) AT = 23a 4  MT =  23a 4         2 T 3I M 36 = (where = 3a) (b) AS = a2  MS = (a2)  2 S I M 16 = (where = 4a) (c) AH = 23 a 6 4         MH =  23 3 a 2  2 H 3I M 24 = (where = 6a) I B I B   Digital Pvt. Ltd. [70] Magnetism (b) Coil or Loop If a loop carries a current of magnitude  and magnetic moment M held in a uniform magnetic field B making an angle  Then the loop experiences a torque given by M B  =  MBsin  =   I(A B) =  If loop have N turns then  NI(A B) =  NIABsin  =  Note: • Torque on dipole is an axial vector and it is directed along axis of rotation of dipole. • Tendency of torque on dipole is try to align the M in the direction of B or tries to makes the axis of dipole parallel to B or makes the plane of coil (or loop) perpendicular to B . • Dipole in uniform magnetic field • Dipole in non-uniform magnetic field Illustration 1: A magnet of magnetic moment ˆ50i A–m2 placed along x–axis. Where magnetic field is B = (0.5 î + 3.0 ĵ ) tesla. The torque acting on magnet is :– Solution: ˆˆ ˆi j k M B 50 0 0 0.5 3 0  =  = = (0) î –(0) ĵ + (150) k̂ N–m = 150 k̂ N–m Illustration 2: A bar magnet of length 10 cm and having the pole strength equal to 10–3 Wb is kept in a magnetic field having magnetic induction (B) equal to 4 × 10–3 T. It makes an angle of 30° with the direction of magnetic induction. The value of the torque acting on the magnet is Solution: Torque,  = MB sin = 0.1 × 10–3 × 4 × 10–3 × sin30° = 10–7 × 4 × 1 2 = 2 × 10–7 Nm Current Carrying Coil MሬሬሬԦ = NIAሬԦ MሬሬሬԦ BሬԦ 𝜃   Digital Pvt. Ltd. [71] Magnetism Illustration 3: A uniform magnetic field of 5000 gauss is established along the positive z-direction. A rectangular loop of side 20 cm and 5 cm carries a current of 10 A is suspended in this magnetic field. What is the torque on the loop in the different cases shown in the following figures ? What is the force in each case ? Which case corresponds to stable equilibrium ? (a) (b) (c) (d) (e) Solution: (a) Torque on loop,  = BIA sin Here,  = 90°; B=5000 gauss = 5000 × 10–4 tesla = 0.5 tesla I = 10 ampere, A = 20 × 5cm2 = 100 × 10–4 = 10–2 m2 Now,  = 0.5 × 10 × 10–2 = 5 × 10–2 Nm It is directed along –y-axis (b) Same as (a). (c)  = 5 × 10–2 Nm along –x-direction (d)  = 5 × 10–2 N m at an angle of 240° with +x direction. (e)  is zero. [ Angle between plane of loop and direction of magnetic field is 90°] Resultant force is zero in each case. Case (e) corresponds to stable equilibrium. Moving Coil Galvanometer Construction   Digital Pvt. Ltd. [72] Magnetism The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis, in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. A spiral spring is also attached to the coil which resists the rotation of coil. Working Principle When a current carrying coil is placed in a magnetic field, it experiences a torque NI(A B) =  𝜏 = 𝑁𝐼𝐴𝐵sin 𝜏 = 𝑁𝐼𝐴𝐵 [in this case I=current through coil, A = Area of coil, B = Magnetic field N = no. of turns,  = 90°] This torque tends to rotate the coil by  angle. The spring S provides a counter torque C that balances the magnetic torque. In equilibrium, C C NIAB I sin NAB  =  =  I   Hence, it means the deflection produced is proportional to the current flowing through the galvanometer. Current Sensitivity Current sensitivity is defined as the deflection produced in the galvanometer when a unit current flows through it. NAB C.S. (rad / A) I C  = = Voltage Sensitivity Voltage sensitivity is equal to the deflection per unit voltage applied across voltmeter. NAB V.S. (rad / V) V IR CR   = = = Advantage of a Moving Coil Galvanometer 1. As the deflection of the coil is proportional to the current passed through it, so a linear scale can be used to measure the deflection. 2. A moving coil galvanometer can be made highly sensitive by increasing N, B, A and decreasing C. 3. As the coil is placed in a strong magnetic field of a powerful magnet, its deflection is not affected by external magnetic fields. This enables us to use the galvanometer in any position. 4. As the coil is wound over a metallic frame, the eddy currents produced in the frame bring the coil to rest quickly. Disadvantage of a Moving Coil Galvanometer 1. The main disadvantage is that its sensitivity cannot be changed at will. 2. All types of moving coil galvanometer are easily damaged by overloading. A current greater than that which the instrument is intended to measure will burn out its hair springs or suspension.   Digital Pvt. Ltd. [75] Magnetism Atomic Magnetism An atomic orbital electron, which is performing bounded uniform circular motion around nucleus. A current constitutes with this orbital motion and hence orbit behaves like current carrying loop. Due to this, magnetic field is produced at nucleus position. This phenomenon called as 'atomic magnetism'. Bohr's postulates : (i) 2mv r = 2 2 kze r (ii) L = mvr = h n 2       , where n = 1, 2,3 ....... Basic elements of atomic magnetism : When electron moves around nucleus in any orbit then a current is induced in this orbit. (a) Effective current I = q T = e T = ef = eω 2π = ev 2πr I = qv 2πr (b) Induced Magnetic Field at center Bin = μoI 2r = μ0e 2Tr = μ0ef 2r = μo 2r ( ev 2πr ) = μoev 4πr2 Bin = μoqv 4πr2 (c) Magnetic Moment M = IA = ( ev 2πr )πr2 = eωr2 2 = evr 2 M = qvr 2 Illustration 1: A helium nucleus is moving in a circular path of radius 0.8m. If it takes 2 sec to complete one revolution. (i) Effective current (ii) Induced magnetic field at the centre (iii) Magnetic moment Solution: (i) Effective current = q I qf T = = {where q = +2e and T = 2s} = 192 1.6 10 2 −  = 19I 1.6 10−=  Amp. (ii) Induced magnetic field at the centre = B = 0q 2Tr  19 0 2 1.6 10 2 2 0.8 −    =   = µ0 × 10–19 T (where q = +2e) (iii) Magnetic moment = M = A = (r2) = ( ) 2191.6 10 0.8−  = 193.21 10− A-m2 e T = 2πr   Digital Pvt. Ltd. [76] Magnetism • Relation between magnetic moment and angular momentum of orbital electron Angular Momentum of electron L = m r Relation between LሬԦ and MሬሬሬԦ M = e r 2  M L = ( e r 2 ) m r = e 2m M L = q 2m for any charge Bohr magneton (µB) According of Bohr's theory, angular momentum of orbital electron is given by L = nh 2 , where n = 1, 2, 3 ........ and h is plank's constant. As per Bohr’s Theory L = nh 2π M L = e 2m M( 2m e ) = nh 2π M = neh 4πm • If n = 1 then M = eh 4 m , which is Bohr magneton denoted by µB • Definition of µB : Bohr magneton can be defined as the magnetic moment of orbital electron which revolves in first orbit of an atom. µB = eh 4 m = 19 34 31 1.6 10 6.6 10 4 3.14 9.1 10 − − −       = 0.923 × 10–23 Am2 Basic elements of atomic magnetism for first orbit of H-atom (n=1, z = 1) (a) Accurate form :- (v = 2.18 × 106 m/sec, f = 6.6 × 1015 cy/sec. r = 0.529Å) • Orbital current  = 0.96 mA • Magnetic induction at nucleus position BN = 12.8 T • Magnetic moment of orbital electron M = 0.923 × 10–23 Am2 (b) Simple form :- (v 2 × 106 m/sec, f 6 × 1015cy/sec, r 0.5Å) • Orbital current  = 1mA • Magnetic induction at nucleus position BN 4T • Magnetic moment of orbital electron M = µB Am2 Illustration 2: A particle of charge 'q' and mass 'm' is moving in a circular orbit with angular speed ''. The ratio of its magnetic moment to that of its angular momentum depends on : Solution: M = qL 2m  M L = q 2m  M L depend on q, m e   Digital Pvt. Ltd. [77] Magnetism Illustration 3: Magnetic moment of orbital electron in first orbit is µB then magnetic moment of that electron in third orbit :- Solution: As we know that, nµ neh 4 m =  So, in 3rd orbit, moment is, 3µ eh 3 4 m =  …(i) and bohr magneton has a value, Bµ eh 4 m =  …(ii) On equating (1), (2), we get µ3 = 3µB Or we can say M  n A nonconducting charged body is rotated with some angular speed :- In this case the ratio of magnetic moment and angular momentum is constant which is equal to q 2m here q = charge and m = the mass of the body. Note : In case of a ring, of mass m, radius R and charge q distributed on it circumference. Angular momentum L = I* = (mR2)() ... (i) [here * = Moment of Inertia] Magnetic moment M = iA = (qf) (R2) M = (q) 2       (R2) = q 2R 2  ...(ii) f = 2   From Equation (i) and (ii) M q L 2m = Although this expression is derived for simple case of a ring, it holds good for other bodies also. For example, for a disc or a sphere. M = qL 2m  ( )q I * M 2m  = , where L = *  Example : R.B. Name Moment of Inertia (*) M qI * 2m =  Ring mR2 21 q R 2  Disc 2mR 2 21 q R 4  Solid sphere 22 mR 5 21 q R 5  sp. shell 22 mR 3 21 q R 3  + + + + + + + + + + + + + + + + + + + + + + + + + + + 𝛚 (q, m, R)   Digital Pvt. Ltd. [80] Magnetism Angle of Dip () • It is the angle which the direction of resultant intensity of earth's magnetic field subtends with horizontal line in magnetic meridian at the given place. • It is the angle made by a freely suspended magnetic needle with the horizontal. • Measured with help of an instrument: Dip Circle Horizontal component of earth's magnetic field (BH) At a given place horizontal component of earth magnetic field is the component of resultant magnetic field of the earth along the horizontal line in magnetic meridian. BH = Bcos and Bv = Bsin .......(1) so that tan  = V H B B and B = 2 2 H V B B+ .......(2) Note : At magnetic poles   = 90° At magnetic equator   = 0° Illustration 1: At a place, the earth's horizontal component of magnetic field is 0.36 × 10–4 Wb/m2. If the angle of dip at that place is 60°, then the vertical component of earth's field at that place in Wb/m2. will be approximately Solution: From the relation, BH = B cos and Bv = B sin v H B B = tan or Bv = BH tan = 0.36 × 10–4 × tan60° = 0.623 × 10–4 Wb/m2 Illustration 2: The earth's magnetic field at a certain place has a horizontal component 0.3 gauss and the total strength 0.5 gauss. The angle of dip is Solution: 2 2 2 2 2 2 2 v H v HB B B B B B (0.5) (0.3) 0.4= +  = − = − = Now, 1v H B 0.4 4 4 tan tan B 0.3 3 3 −    = = =  =     Illustration 3: If the angles of dip at two places are 30° and 45° respectively, then the ratio of horizontal components of earth's magnetic field at the two places will be Solution: The horizontal components are (BH)1 = B cos1 and (BH)2 = B cos2  H 1 1 H 2 2 (B ) cos cos30 3 3 2 (B ) cos cos45 2 2   = = =  =   𝑩𝑬 𝜽 Magnetic Meridian   Digital Pvt. Ltd. [81] Magnetism Aura Effect Due to cosmic particles near Poles. Illustration 4. Density of cosmic particles (positive charge) is more ? (1) Poles (2) Equator Solution. Density of cosmic particles (positive charge) is more at poles. The earth's magnetic field is strongest at the poles hence cosmic charges particles are deflected away from the equator. Angle of Declination and Apparent Angle of Dip Angle Declination(𝛅) At a given place the acute angle between geographic meridian and the magnetic meridian is called angle of declination, i.e. at a given place it is the angle between the geographical north-south direction and the direction indicated by a magnetic compass needle in its equilibrium. Apparent Angle of Dip (𝛉′) • When Magnetic needle is free to rotate only in a vertical plane other than the magnetic meridian, its angle with the horizontal is different than actual angle of dip. • If the dip circle is not kept in the magnetic meridian, the needle will not show the correct direction of earth's magnetic field. • The angle made by the needle with horizontal is called the apparent dip for this plane. Geographical North 𝛿   Digital Pvt. Ltd. [82] Magnetism If dip circle is at an angle  to the meridian, the effective horizontal component in this plane is BHcos. Here, BV = BE sin θ B′H = BH cos α = BE cos θ cos α tan θ′ = BV B′H = BE sin θ BE cos θ cos α tan θ′ = tanθ cosα …(i) Apparent dip angle is always greater than actual dip angle Now if the dip circle is rotated through an angle 90° for this position. It will make angle (90 - ) with the meridian, lets this time apparent dip is 'a Thus a tan tan tan ' cos(90 ) sin    = = −   ...(2) Note : as sin2 + cos2 = 1  cot2a + cot2'a = cot2 Thus, one can get the true dip angle , without locating the magnetic meridian. If apparent dip angles in two vertical planes which are at 90º from each other are 1 and 2 then actual dip angle  is given by :- cot2 θ = cot2 θ1 + cot2 θ2 Illustration 1: A compass needle whose magnetic moment is 60 A m2 pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is 40µ Wb/m2, experiences a torque 1.2 × 10–3 Nm. What is the declination at this place? Solution: As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian, so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal component of earth's magnetic field, i.e.,  = MBHsin where  = angle between geographical and magnetic meridians called angle of declination So, 3 6 1.2 10 1 sin 30 60 40 10 2 − −   = =   =    𝛼 M ag n et ic M er id ia n 𝑩𝑬 𝜃′ 𝑩𝑯 𝑩𝑽 𝑩′𝑯 𝑩′𝑬 𝛼 M ag n et ic M er id ia n 𝑩𝑬 𝜃′ Geographical North 𝛿 mBH mBH Magnetic meridian   Digital Pvt. Ltd. [85] Magnetism (ii) From  = Ktan , 1 tan K   ( = same in series combination) KA > KB tanA < tanB  A < B Vibration Magnetometer It is an instrument used to compare the horizontal component of magnetic field of earth of two different places, to compare magnetic fields and magnetic moments of two bar magnets. It is also called oscillation magnetometer. Working Principle : Angular SHM This device works on the principle, that whenever a freely suspended bar magnet in earth horizontal magnetic field component (BH) is slightly disturbed from its equilibrium position then, it will experience a torque and execute angular S.H.M. (Rotation is possible only in horizontal plane.) When a dipole is suspended in a uniform magnetic field it will align itself parallel to field. Now if it is given a small angular displacement  with respect to its equilibrium position. The restoring torque acts on it :- τ = −MBH sinθ For small angle, sinθ ≈ θ τ = −MBHθ For angular SHM, τ = −Iω2θ ⇒ −Iω2θ = −MBHθ ⇒ ω = √ MBH I ⇒ 2π T = √ MBH I ⇒ T = 2π√ I MBH where : M : magnetic moment of bar magnet  : Moment of Inertia = ml 2 12 {m : mass of magnet, : length of magnet} Comparison of two magnets of Same Size (Same moment of inertia) Let the two magnets of same size have moment of inertia I and magnetic moments M1 and M2. Suspend these two given magnets one by one in the metal stirrup of the vibration magnetometer and note the time period in each case. 1 1 H T 2 M B  =  …(i) 2 2 H I T 2 M B =  …(ii) 𝑩𝑯 𝐁𝑯 𝐌𝟏 𝐁𝐇 𝐌𝟐   Digital Pvt. Ltd. [86] Magnetism Dividing equation (i) and (ii) 1 2 2 1 T M T M = or 2 1 2 2 2 1 M T M T = By knowing T1 and T2 the ratio 1 2 M M can be determined. Comparison of earth's magnetic field at two different places (different BH) Let the vibrating magnet have moment of inertia I and magnetic moment M. Let it be vibrated at places, where earth's magnetic field is 1HB and 2HB Then, T1 = 2 1H I MB and T2 = 2 2H I MB T1 and T2 are determined by placing magnetometer at two different places, one by one. Dividing, 2 1 H1 2 H BT T B = 2 1 2 2 H1 1 2 2 2 2 H 1 12 2 BT T B cos B B cosT T   =  =  2 1 2 2 2 2 1 1 B T cos B T cos  =  By knowing T1, T2 and 1, 2 the ratio 1 2 B B can be determined. Combination of Two magnets Let the two magnets have moments of inertia 1 and 2 and magnetic moments M1 and M2 respectively. Place these two given magnets one upon the other as shown in Figure (a). This combination is called sum combination. This combination in the magnetometer and set it into oscillations. The time period of combination T1 is determined. Inet = I1 + I2 Mnet = M1 +M2 T1 = 2π√ I1+I2 (M1+M2)BH …(i) Now, the two magnets are placed as shown in Figure (b). This combination is called 'difference combination'. This combination is put in the magnetometer and its time period T2 is determined. Inet = I1 + I2 Mnet = M1 −M2 T2 = 2π√ I1+I2 (M1−M2)BH …(ii) Dividing, 1 1 2 2 1 2 T M M T M M − = + [from equation (i) and (ii)] 𝐁𝐇𝟏 𝐌 𝐁𝐇𝟐 𝐌 Sum combination difference combination   Digital Pvt. Ltd. [87] Magnetism Note : If we cut any bar magnet into n equal parts normal to its length, then time period of vibration in external magnetic field- T′ = T n If we cut any bar magnet into n equal parts along its length, then time period of vibration in external magnetic field- T′ = T Illustration 1: Time period of thin bar magnet of vibration magnetometer is 'T'. If it cuts into two equal parts in then the time period of each part fashions. (a) Along or parallel to its length (b) Perpendicular to its length, Solution: Time period of thin rectangular bar magnet T = 2 H I MB Where  = 2ml 12 T  I M ........ (1) [BH → same] Case I Cutted parallel to its length :- for each part T´  I / 2 M / 2 ........(2) from (1) and (2) Time period of each part T´ = T Special note :- If bar magnet cuts into 'n' equal parts parallel to its length then time period of each part becomes. T´ = T Case - II Perpendicular to its length :- for each part T´  I / 8 M / 2 ......... (3)   Digital Pvt. Ltd. [90] Magnetism Baxis = BH  0 4   3 2M x = BH (If x >>> ) Illustration 1: A short bar magnet placed on the table along the north south line. Its north pole is directed towards geographic north then location of neutral points is :- Solution: When N–pole of magnet directed towards North. Two neutral points symmetrically located on equatorial line of magnet. Location of neutral point is East and West Illustration 2: The ratio of the magnetic field due to a small bar magnet in end on position to broad side on position is :- Solution: 3 axis 3 eq. B 2KM / d 2 B 1KM / d Illustration 3: A short magnet of moment 6.75 A–m2 produces neutral points on its axis. If the horizontal component of earth's magnetic field 5 x 10–5 Wb/m2, then the distance of the neutral points from the centre of magnet:- Solution: BH = 3 2KM d d = 1/3 1/37 5 H 2KM 2 10 6.75 B 5 10 − −      =       = 0.3 m. = 30 cm. Illustration 4: The earth's magnetic field at the equator is approximately 0.4 G. Estimate the earth's dipole moment. Solution: Equatorial magnetic field BE = 0 3 M 4 r   Magnetic moment M = 3 E 0 B (4 r )    Digital Pvt. Ltd. [91] Magnetism where BE = 0.4G = 4 × 10–5T, r = 6.4 × 106 m = 5 6 3 0 4 10 (6.4 10 ) / 4 −     = 4 × 102 × (6.4 × 106)3 = 1.04 × 1023 A–m2 Magnetization Magnetization, also termed as magnetic polarization, is a vector quantity that gives the measure of the density of permanent or induced dipole moment in a given magnetic material. The magnetic effects of a material can also be induced by passing an electrical current through the material; the magnetic effect is caused by the motion of electrons in atoms, or the spin of the electrons or the nuclei. Magnetizing field or Magnetic Intensity (𝐇ሬሬԦ) Field in which a material is placed for magnetization, called as magnetizing field. HሬሬԦ = BሬԦo μo unit of H : A/m BሬԦo: Magnetic Field μo: Permeability of free space Intensity of magnetization (𝐈Ԧ) When a magnetic material is placed in magnetising field then induced dipole moment per unit volume of that material is known as intensity of magnetisation. IԦ = MሬሬሬԦ V unit of I : A/m 2 3 M IA ampere meter V V meter    = =     Induced magnetic field due to these induced dipoles in the material is given by: BሬԦin = μoIԦ 𝐻ሬԦ = 𝐵ሬԦ𝑜 𝜇𝑜 𝐵ሬԦ𝑜 𝐻ሬԦ = 𝐵ሬԦ𝑜 𝜇𝑜 𝐵ሬԦ𝑜   Digital Pvt. Ltd. [92] Magnetism Magnetic susceptibility ( 𝛘𝐦 ) m = I H [It is a scalar with no units & dimensions] Physically it represent the ease with which a magnetic material can be magnetised Note:- A material with more m , can be change into magnet easily. Magnetic Permeability ( 𝛍 ) μ = B H = Total magnetic field inside a material Magnetizing field Unit of µ : µ = 2 mB Wb/ m Weber H A H H A / m A m A m m −  = = = − − It represents the degree up to which a material can be penetrated by the magnetic field lines Relative Permeability (µr) r 0 µ µ µ = It has no units and dimensions. Relation between 𝛍 and 𝛘𝐦 When a magnetic material is placed in a magnetic field BሬԦo for magnetization, then total magnetic field in the material is BሬԦm = BሬԦo + BሬԦin {BሬԦin= induced field.} 0B = µ0 H ; inB = µ0 I  BሬԦm = μo(HሬሬԦ + IԦ) Bm = μo(H + I) μH = μoH(1 + I H ) μ = μo(1 + χm) μ μo = (1 + χm) μr = (1 + χm) For vacuum, μr = 1 , χm = 0 For air, μr = 1.04 , χm = 0.04 Illustration 1: A solenoid of 200 turns/m is carrying a current of 5A. Relative permeability of core material of solenoid is 3000. Determine the magnitudes of the magnetic intensity, magnetization and the magnetic field inside the core. Solution: magnetic intensity H = ni = 200 m–1 × 5A = 1000 Am–1 µr = 1 + , i.e.,  = 3000 – 1 = 2999 𝐻ሬԦ 𝐼Ԧ = 𝑀ሬሬԦ 𝑉 𝐻ሬԦ 𝐼Ԧ   Digital Pvt. Ltd. [95] Magnetism which are randomly oriented. (i.e. in absence of external magnetic field the magnetic moment of whole material is zero) When rod of material is suspended between poles of magnet. It becomes perpendicular to the direction of external magnetic field. If there is strong magnetic field in between the poles then rod becomes parallel to the magnetic field. Weak magnetic field between magnetic poles can made rod parallel to field direction. Magnetic moment of substance in presence of external magnetic field Value M is very less and opposite to H . Value M is low but in direction of H . M is very high and in direction of H . Examples Bi, Cu, Ag, Pb, H2O, Hg, H2, He, Ne, Au, Zn, Sb, NaCl, Diamond.( May be found in solid, liquid or gas). Na, K, Mg, Mn, Sn, Pt, Al, O2 (May be found in solid, liquid or gas.) Fe,Co, Ni all their alloys, Fe3O4 Gd, Alnico, etc. (Normally found only in solids) (crystalline solids) Note : Above Curie temperature, ferromagnetic materials lose their permanent magnetic properties and behave as paramagnetic materials. Illustration 1: A uniform magnetic field gets modified as shown below, when the specimens X and Y are placed in it. (a) Identify the two specimens X and Y (b) State the reason for the behaviour of the field lines in X and Y.   Digital Pvt. Ltd. [96] Magnetism Solution: (a) X → Diamagnetic Y → Paramagnetic (b) Magnetic permeability of diamagnetic specimen is slightly less than vacuum and magnetic permeability of paramagnetic specimen is slightly more than vacuum. Illustration 2: If a ferromagnetic material is inserted in a current carrying solenoid, the magnetic field of solenoid Solution: The flux inside the solenoid increases due to the presence of ferromagnetic material and hence the field of solenoid increases largely. Illustration 3: Susceptibility of Mg at 300 K is 1.2 × 10–5. The temperature at which susceptibility will be 1.8 × 10–5 is Solution: 1 T    1T1 = 2T2 Hence, 5 2 5 1.2 10 300 T 200K 1.8 10 − −   = =  Magnetic Hysteresis Only Ferromagnetic materials show magnetic hysteresis, when Ferromagnetic material is placed in external magnetic field for magnetization. [Lagging of B behind H is called hysteresis] • When magnetizing field is increased across a ferromagnetic material • The magnetic field inside the material also increases • When this magnetizing field is removed, • There is some residual magnetic field in the material known as Retentivity • To remove this residual magnetic field inside the material, • external magnetizing field is again applied but in opposite direction till net magnetic field inside the material becomes zero. • This magnetizing field required to destroy residual magnetic field is known as Coercivity. B H O A B Br B H O A B C HC   Digital Pvt. Ltd. [97] Magnetism • As external magnetizing field is further increased, a magnetic field is also developed in the material. • Magnetizing field again becomes zero and then it is again applied in forward direction. • Due to lagging behind of B with H this curve is known as hysteresis curve. [Lagging of B behind H is called hysteresis] Cause of hysteresis : By removing external magnetizing field (H = 0), the magnetic moment of some domains remains aligned in the applied direction of previous magnetizing field which results into a residual magnetism. Residual magnetism (OB) = Br (retentivity/remanence) : Retentivity of a specimen is a measure of the magnetic field remaining in the ferromagnetic specimen when the magnetizing field is removed. Coercivity (OC) = Hc : Coercivity is a measure of magnetizing field required to destroy the residual magnetism of the ferromagnetic specimen. Hysteresis Loss (i) The area of hysteresis loop is equal to the energy loss per cycle per unit volume. (ii) Its value is different for different materials. (iii) The work done per cycle per unit volume of material is equal to the area of hysteresis loop.  Total energy loss in material = WH  i.e WH = volume of material × area of hysteresis curve × frequency × time. WH = V × A × n × t joule WH = V × A × n × t J calorie Soft and Hard Magnetic Materials Soft magnetic materials Hard magnetic materials 1. Low retentivity 2. Low coercivity 3. Small hysteresis Loss 4. Suitable for making electromagnets, cores of transformers etc. e.g. Soft Iron 1. High retentivity 2. High coercivity 3. Large hysteresis Loss 4. suitable for permanent magnet e.g. Steel, Alnico B H B H B H O A B C D E F