Magnetic Fields Due to Currents: Lecture 36, Study notes of Physics

Download Magnetic Fields Due to Currents: Lecture 36 and more Study notes Physics in PDF only on Docsity! 1 Lecture 36 – Magnetic fields due to currents Chapter # 29 Magnetic field due to a current Force between tow parallel currents Ampere’s law Solenoids and toroids A current carrying coil as a magnetic dipole 2 Our Study of Magnetism • Lorentz Force Eqn • Motion in a uniform B-field BvqEqF rrrr ×+= circular orbit • Forces on charges moving in wires qB mvrc = BlIdFd rrr ×= • Magnetic dipole IA rr =μ B rrr ×= μτ BU rr ⋅−= μ • Today: fundamentals of how currents generate magnetic fields 5 Calculating the Magnetic Field due to a Current • Similarly, the magnetic field )var( 4 /1026.1 )(/104 sin 4 3 0 6 7 0 2 0 lawtSaandBiot r rsdiBd AmT typermeabilimagneticAmT r dsidB rrr × = ⋅×≈ ⋅×= = − − π μ πμ θ π μ B-field of a Straight Wire • Calculate field at point P using Biot-Savart Law: • Rewrite in terms of R,θ : x Rr θ θ P Idx 3 0 sin)( 4 r θrdx π IμdBB ∫∫ +∞ ∞− == sin Rr θ = ⇒ x Rθ − =tan ⇒ θRx cot−= therefore, dθ θ Rdx ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −−= 2sin 1 ⇒ R θdθ r θdx sin)(sin 2 = 3 0 4 r rxd π IμBd rrr × = Which way is B? +z y ∫= π θdθ πR IμB 0 0 sin 4 ⇒ [ ] πθ πR IμB 00 cos4 −= therefore, πR IμB 2 0= B-field of a Straight Wire x Rr θ θ P Idx θ R dθ π IμB π sin 40 0∫= 10 Checkpoint 29-1 The figure shows three circles consisting of concentric circular arcs (either half- or quarter-circles of radii r, 2r, and 3r) and radial lengths. The circuits carry the same current. Rank them according to the magnitude of the magnetic field produced at the center of the curvature (the dot), greatest first. R iB π φμ 4 0= 3 4 4)3(4)(4 )( 000 r i r i r iBBBa ud μ π πμ π πμ =+=+= 3 2 4)3(4)(4 )( 000 r i r i r iBBBb ud μ π πμ π πμ =−=−= 12 13 43 1 2 1 4 1 4)3(4)(4 )2/( )2(4 )2/()( 00000 r i r i r i r i r iBBBBc udrdl μμ π πμ π πμ π πμ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++=++=++= )(),(),( bca 11 Sample Problem 29-2 Figure 30-8a shows two long parallel wires carrying currents i1 and i2 in opposite directions. What are the magnitude and direction of the net magnetic field at point P? Assume: i1 = 15 A, i2 = 32 A, and d = 5.3 cm. R iB R iB π μ π μ 22 20 2 10 1 == R iB π μ 2 0= oo dR d R 45cos45cos =⇒= oo d iB d iB 45cos245cos2 20 2 10 1 π μ π μ == T m AAAmT ii d BBB o o 4 2 227 2 2 2 1 02 2 2 1 1089.1 45cos)103.5)(2( )32()15()/104( 45cos2 − − − ×= × +⋅× = +=+= π π π μ 2 1tan B B =φ o A A i i B B 25 32 15tantantan 1 2 11 2 11 ==== −−−φ axisxtrwoo −=+= ...7045φθ Putting it all together • We know that a current-carrying wire can experience force from a B-field. • We know that a a current-carrying wire produces a B- field. • Therefore: We expect one current-carrying wire to exert a force on another current-carrying wire: • Current goes together wires come together • Current goes opposite wires go opposite d Ia Ib F r F r 15 Checkpoint 29-2 The figure shows three long, straight, parallel, equally spaced wires with identical currents either in or out of the page. Rank the wires according to the magnitude of the force on each due to the currents in the other two wires. d iLiF baba π μ 2 0=abF r acF r caF r cbF r iiii cba === 2 1 2)2(22 2 0 2 0 2 0 d Li d Li d LiFFF acaba π μ π μ π μ =−=−= 2 222 2 0 2 0 2 0 d Li d Li d LiFFF bcbab π μ π μ π μ =+=+= baF r bcF r 2 3 22)2(2 2 0 2 0 2 0 d Li d Li d LiFFF cbcac π μ π μ π μ =+=+= acb FFF >> Example • A current I flows in the +y direction in an infinite wire; a current I also flows in the loop as shown in the diagram. – What is Fx, the net force on the loop in the x-direction? (a) Fx < 0 (b) Fx = 0 (c) Fx > 0 Fleft Fright • You may remember that the net force on a current loop in a uniform B- field is zero, but the B-field produced by an infinite wire is not uniform! • Forces cancel on the top and bottom of the loop. • Forces do not cancel on the left and right sides of the loop. • The left segment is in a larger magnetic field than the right • Therefore, Fleft > Fright Ftop Fbottom X I I x y Example (cont) R I Two nice things about calculating B at the center of the loop: 0 3 ( ) 4π μ I dl RB dB R = = But how do we calculate B? We must use Biot-Savart Law to calculate the magnetic field at the center: 0 34 I dl rdB r μ π × = r rr • I dl is always perpendicular to r: r into page • r is constant (r = R) dl r× = r r 0 24π μ I dl R = 0 2 (2π )4π μ I R R = 0 2 μ I R = Amp meter B ∝Note: rr dl r Circular Loop Bz = μ0 iR2 2(z 2+R2)3/2 R B z z 0 0 ≈ 1z 3 Example (a) Bz(A) < 0 (b) Bz(A) = 0 (c) Bz(A) > 0 3A • The right current loop gives rise to Bz <0 at point A. • The left current loop gives rise to Bz >0 at point A. • From symmetry, the magnitudes of the fields must be equal. • Therefore, B(A) = 0 x o x o z I I A B • Equal currents I flow in identical circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. – What is the magnetic field Bz(A) at point A, the midpoint between the two loops? Example • Equal currents I flow in identical circular loops as shown in the diagram. The loop on the right (left) carries current in the ccw (cw) direction as seen looking along the +z direction. – What is the magnetic field Bz(B) at point B, just to the right of the right loop? 3B (a) Bz(B) < 0 (b) Bz(B) = 0 (c) Bz(B) > 0 x o x o z I I A B • The signs of the fields from each loop are the same at B as they are at A! • However, point B is closer to the right loop, so its field wins!