Faraday's Law of Induction: Induced Electric Fields and Emf, Study notes of Physics

Download Faraday's Law of Induction: Induced Electric Fields and Emf and more Study notes Physics in PDF only on Docsity! Faraday's Law dS B B ΦB B dS≡ •∫ ε = − d dt BΦ Global Review • Electrostatics » motion of “q” in external E-field » E-field generated by Σqi • Magnetostatics » motion of “q” and “i” in external B-field » B-field generated by “I” • Electrodynamics » time dependent B-field generates E-field ac circuits, inductors, transformers, etc » time dependent E-field generates B-field • electromagnetic radiation - light Induction Effects from Currents • Switch closed (or opened) ⇒ current induced in coil b • Steady state current in coil a ⇒ no current induced in coil b a b • Conclusion: A current is induced in a loop when: • there is a change in magnetic field through it • this can happen many different ways • How can we quantify this? An example of induction downward velocity A wire loop falling into an increasing magnetic field time magnetic field Force acting on moving charges N N N Faraday's Law • Define the flux of the magnetic field through an open surface as: dS B BΦB B dS≡ •∫ • Faraday's Law: The emf induced in a circuit is determined by the time rate of change of the magnetic flux through that circuit. The minus sign indicates direction of induced current (given by Lenz's Law). ε = − d dt BΦ So what is this emf?? ACT 1 • A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. – What is the direction of the induced current in the loop?1A X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X v x y (a) ccw (b) cw (c) no induced current • A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown. • What is the direction of the induced current in the loop? (a) ccw (b) cw (c) no induced current 1B v I x y ACT 1 • A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. – What is the direction of the induced current in the loop? (c) no induced current(a) ccw (b) cw 1A X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X v x y • There is a non-zero flux ΦB passing through the loop since B is perpendicular to the area of the loop. • Since the velocity of the loop and the magnetic field are CONSTANT, however, this flux DOES NOT CHANGE IN TIME. • Therefore, there is NO emf induced in the loop; NO current will flow!! ACT 1 • A conducting rectangular loop moves with constant velocity v in the +x direction through a region of constant magnetic field B in the -z direction as shown. – What is the direction of the induced current in the loop? y X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X v x1A • The flux through this loop DOES change in time since the loop is moving from a region of higher magnetic field to a region of lower field. • Therefore, by Lenz’ Law, an emf will be induced which will oppose the change of flux. • The current i is induced in the clockwise direction to restore the flux. (a) ccw (b) cw (c) no induced current • A conducting rectangular loop moves with constant velocity v in the -y direction and a constant current I flows in the +x direction as shown. • What is the direction of the induced current in the loop?1B (a) ccw (b) cw (c) no induced current v I x y i ∆B/∆t → E • Faraday's law ⇒ a changing B induces an emf which can produce a current in a loop. • In order for charges to move (i.e., the current) there must be an electric field. ∴ we can state Faraday's law more generally in terms of the E field which is produced by a changing B field. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x r E E E E B • Suppose B is increasing into the screen as shown above. An E field is induced in the direction shown. To move a charge q around the circle would require an amount of work = • This work can also be calculated from ε = W/q. W qE dl= •∫ ∆B/∆t → E • Putting these 2 eqns together: ⇒ • Therefore, Faraday's law can be rewritten in terms of the fields as: x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x r E E E E B W qE dl= •∫ ε = •∫ E dl ε = W q E dl d dt B•∫ = − Φ 3 • Note that for E fields generated by charges at rest (electrostatics) since this would correspond to the potential difference between a point and itself. Consequently, there can be no "potential function" corresponding to these induced E fields. E dl•∫ = 0 Rate of change of flux through loop Line integral around loop ACT 2 • The magnetic field in a region of space of radius 2R is aligned with the z-direction and changes in time as shown in the plot. – What is sign of the induced emf in a ring of radius R at time t=t1? 3A 3B – What is the relation between the magnitudes of the induced electric fields ER at radius R and E2R at radius 2R ? t Bz t1 X X X X X X X X X X X X X X X X X X X X X X X X X X X X x y X X X X X X X X X X X X X X X X X X X X X X X X X X X X R (a) ε < 0 ( E ccw) (b) ε = 0 (c) ε > 0 ( E cw) (a) E2R = ER (b) E2R = 2ER (c) E2R = 4ER Demo E-M Cannon v • Connect solenoid to a source of alternating voltage. ~ side view • • F F •B B B top view2 • The flux through the area ⊥ to axis of solenoid therefore changes in time. • A conducting ring placed on top of the solenoid will have a current induced in it opposing this change. • There will then be a force on the ring since it contains a current which is circulating in the presence of a magnetic field. ACT 3 • For this ACT, we will predict the results of variants of the electromagnetic cannon demo . – Suppose two aluminum rings are used in the demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2. 2B – Suppose two identically shaped rings are used in the demo. Ring 1 is made of copper (resistivity = 1.7X10-8 Ω-m); Ring 2 is made of aluminum (resistivity = 2.8X10-8 Ω-m). Let F1 be the force on Ring 1; F2 be the force on Ring 2. (a) F2 < F1 (b) F2 = F1 (c) F2 > F1 2A Ring 1 Ring 2 (a) F2 < F1 (b) F2 = F1 (c) F2 > F1 ACT 3 Ring 1 Ring 2 • For this ACT, we will predict the results of variants of the electromagnetic cannon demo which you just observed. – Suppose two aluminum rings are used in the demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2. 2A (a) F2 < F1 (c) F2 > F1(b) F2 = F1 • The key here is to realize exactly how the force on the ring is produced. • A force is exerted on the ring because a current is flowing in the ring and the ring is located in a magnetic field with a component perpendicular to the current. • An emf is induced in Ring 2 equal to that of Ring 1, but NO CURRENT is induced in Ring 2 because of the slit! • Therefore, there is NO force on Ring 2!