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Faraday’s Law and Lenz’s Law B(t) i ~ Physics 1304: Lecture 13, Pg 1 Overview of Lecture Induction Effects Faraday’s Law (Lenz’ Law) • Energy Conservation with induced currents? Faraday’s Law in terms of Electric Fields Text Reference: Chapter 31.1-4 Physics 1304: Lecture 13, Pg 2 Induction Effects Bar magnet moves through coil S N N S N S v Current induced in coil • Change pole that enters Induced current changes sign Bar magnet stationary inside coil No current induced in coil Coil moves past fixed bar magnet Current induced in coil v v S N Physics 1304: Lecture 13, Pg 3 Induction Effects from Currents • a Switch closed (or opened) b current induced in coil b • Steady state current in coil a no current induced in coil b Conclusion: A current is induced in a loop when: • there is a change in magnetic field through it • loop moves through a magnetic field How can we quantify this? Physics 1304: Lecture 13, Pg 4 Faraday's Law • Define the flux of the magnetic field through a surface (closed or open) from: dS B B 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). Physics 1304: Lecture 13, Pg 5 Lenz's Law • Lenz's Law: The induced current will appear in such a direction that it opposes the change in flux that produced it. B B S N v N S v Conservation of energy considerations: Claim: Direction of induced current must be so as to oppose the change; otherwise conservation of energy would be violated. » Why??? If current reinforced the change, then the change would get bigger and that would in turn induce a larger current which would increase the change, etc.. Physics 1304: Lecture 13, Pg 6 Lecture 18, CQ y 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 1A the loop? (a) ccw (b) cw (b) cw x (c) no induced current • A conducting rectangular loop moves with y 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 1B current in the loop? (a) ccw XXXXXXXXXXXX XXXXXXXXXXXX X X X X X X X vX X X X X XXXXXXXXXXXX I v x (c) no induced current Physics 1304: Lecture 13, Pg 7 Lecture 16, ACT 1 A conducting rectangular loop moves with y 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 1A the loop? (a) ccw (b) cw XXXXXXXXXXXX XXXXXXXXXXXX X X X X X X X vX X X X X XXXXXXXXXXXX x (c) no induced current • There is a non-zero flux FB 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!! Physics 1304: Lecture 13, Pg 8 Lecture 16, ACT 1 y 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? (a) ccw (b) cw (b) cw x (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 XXXXXXXXXXXX XXXXXXXXXXXX X X X X X X X vX X X X X XXXXXXXXXXXX y I i v x (c) no induced current • 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. Physics 1304: Lecture 13, Pg 9 Demo E-M Cannon v Connect solenoid to a source of alternating voltage. The flux through the area ^ to axis of solenoid therefore changes in time. ~ side view A conducting ring placed on top of the solenoid will have a current induced in it opposing this change. F B There will then be a force on the ring since it contains a current which is circulating in the presence of a magnetic field. B F B top view Physics 1304: Lecture 13, Pg 10 Lecture 18, CQ Let us predict the results of variants of the electromagnetic cannon demo which you just observed. Ring 1 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. Ring 2 (a) F2 < F1 (b) F2 = F1 (c) F2 > F1 – Suppose two identically shaped rings are used in the demo. Ring 1 is made of copper (resistivity = 1.7X10-8 W-m); Ring 2 is made of aluminum (resistivity = 2.8X10-8 W-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 Physics 1304: Lecture 13, Pg 11 Lecture 18, CQ For this ACT, we will predict the results of variants of the electromagnetic cannon demo which you just observed. Ring 1 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. Ring 2 (a) F2 < F1 (b) F2 = F1 (c) 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! Physics 1304: Lecture 13, Pg 12 Lecture 18, CQ For this ACT, we will predict the results of variants of the electromagnetic cannon demo which you just observed. Ring 1 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. Ring 2 (a) F2 < F1 (b) F2 = F1 (c) F2 > F1 – Suppose two identically shaped rings are used in the demo. Ring 1 is made of copper (resistivity = 1.7X10-8 W-m); Ring 2 is made of aluminum (resistivity = 2.8X10-8 W-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 • The emf’s induced in each ring are equal. • The currents induced in each ring are NOT equal because of the different resistivities. • The copper ring will have a larger current induced (smaller resistance) and therefore will experience a larger force (F proportional current). Physics 1304:to Lecture 13, Pg 13 Calculation Suppose we pull with velocity v a coil of resistance R through a region of constant magnetic field B. What will be the induced current? » What direction? xxxxxx I xxxxxx xxxxxx w Lenz’ Law clockwise!! x x x x x x x What is the magnitude? » Magnetic Flux: » Faraday’s Law: \ Physics 1304: Lecture 13, Pg 14 v Energy Conservation? • The induced current gives rise to a net magnetic force ( F ) on the loop which opposes the motion. F' xxxxxx F • Agent must exert equal but I xxxxxx xxxxxx w xxxxxx F' x opposite force to move the loop with velocity v; \ agent does work at rate P, where • Energy is dissipated in circuit at rate P' P = P' ! Physics 1304: Lecture 13, Pg 15 v Faraday's Law • Define the flux of the magnetic field through a surface (closed or open) from: dS B B 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). Physics 1304: Lecture 13, Pg 16 Conceptual Question In figure (a), a solenoid produces a magnetic field whose strength increases into the plane of the page. An induced emf is established in a conducting loop surrounding the solenoid, and this emf lights bulbs A and B. In figure (b), points P and Q are shorted. After the short is inserted, 1. 2. 3. 4. 5. 6. bulb A goes out; bulb bulb B goes out; bulb bulb A goes out; bulb bulb B goes out; bulb both bulbs go out. none of the above B A B A gets gets gets gets brighter. brighter. dimmer. dimmer. Physics 1304: Lecture 13, Pg 17 Physics 1304: Lecture 13, Pg 18 DB E • Ex x x x x x x x x x Faraday's law a changing B induces an E emf which can produce a current in a loop. xxxxxxxxxx In order for charges to move (i.e., the r xxxxxxxxxx current) there must be an electric field. \ we can state Faraday's law more generally x x x xBx x x x x x in terms of the E field which is produced by E a changing B field. x x x x x x x xEx x 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 = W qE dl • This work can also be calculated from e = W/q. Physics 1304: Lecture 13, Pg 19 DB E • Putting these 2 eqns together: • Therefore, Faraday's law can be rewritten in terms of the fields as: x x xEx x x x x x x E xxxxxxxxxx r xxxxxxxxxx B xxxxxxxxxx E x x x x x x x xEx x 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. Physics 1304: Lecture 13, Pg 20 Lecture 18, CQ 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? (a) e < 0 ( E ccw) (b) e = 0 (c) e > 0 ( E cw) – What is the relation between the magnitudes of the induced electric fields ER at radius R and E2R at radius 2R ? (a) E2R = ER (b) E2R = 2ER y XXXX XXXXXXX XXXXXXXX XXXXXXXXX R XXXXXXXXX XXXXXXXX XXXXXXX XXXX Bz t1 x t (c) E2R = 4ER Physics 1304: Lecture 13, Pg 21 y Lecture 18, CQ 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? (a) e < 0 ( E ccw) (b) e = 0 (c) e > 0 ( E cw) • There will be an induced emf at t=t1 because the magnetic field (and therefore the magnetic flux) is changing. It makes NO DIFFERENCE that at t=t1 the magnetic field happens to be equal to ZERO! • The magnetic field is increasing at t=t1 (actually at all times shown!) which induces an emf which opposes the corresponding change in flux. ie electric field must be induced in a counter clockwise sense so that the current it would drive would create a magnetic field in the z direction. XXXX XXXXXXX XXXXXXXX XXXXXXXXX R XXXXXXXXX XXXXXXXX XXXXXXX XXXX x B t1 t Physics 1304: Lecture 13, Pg 22 y Lecture 18, CQ 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? (a) e < 0 ( E ccw) (b) e = 0 (c) e > 0 ( E cw) XXXX XXXXXXX XXXXXXXX XXXXXXXXX R XXXXXXXXX XXXXXXXX XXXXXXX XXXX B What is the relation between the magnitudes of the induced electric fields ER at radius R and E2R at radius 2R ? (a) E2R = ER (b) E2R = 2ER t1 x t (c) E2R = 4ER • The rate of change of the flux is proportional to the area: • The path integral of the induced electric field is proportional to the radius. NOTE: This result is important for the operation of the Betatron. Physics 1304: Lecture 13, Pg 23 Textbook Problems Problems 31-33, 31-39, 31-46, 31-67 Physics 1304: Lecture 13, Pg 24 Induced EMF by motion Physics 1304: Lecture 13, Pg 25 Rod Moving in B-Field Consider a metal rod moving in a B-field. The free charges in the rod will experience a force given X X X X X X X X X X X X - - -by, X X X X X X X X - - F qv B the top the force would be upwards and negative charge would thus accumulate on end. 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 The work done in separating the charges is: W F l qvBl By definition the EMF is then, e W q i.e. e vBl Physics 1304: Lecture 13, Pg 26 The previous result is connected with Faraday’s law. To see this we re-interpret Faradays law as the induced EMF along the path of a moving conductor in the presence of a constant of changing B-field. This induced EMF would be equal to the rate at which magnetic flux sweeps across the path. x x x x x x x x x x x x x x x x x x x x x x x x Ds dFB dt e vBL Ds BL Dt Aswept DsL e d d (sBL) ( BA) dt dt e dFB dt : rate at which flux is swept. Physics 1304: Lecture 13, Pg 27 Electric Generators and Motors n̂ Consider a loop of wire in a constant magnetic field. If we rotate the loop the flux through the cross sectional area will change. By Faradays Law an Induced EMF will be generated. B FB BA e B i i FB (t ) BA cos (t ) i The converse is also true if we run a current through the wire then a torque will be excerted on the loop making it turn. Physics 1304: Lecture 13, Pg 28 Textbook Problems Physics 1304: Lecture 13, Pg 29 Textbook Problems Physics 1304: Lecture 13, Pg 30 Electric Motors and Generators Physics 1304: Lecture 13, Pg 31 Textbook Problems Physics 1304: Lecture 13, Pg 32