hysics 40 Homework olutions - Walker Chapter 3 Conceptual Questions CQ5. Before the switch is closed there is no current in the coil and therefore no netic flux through the metal ring. When the switch is closed there begins to be a netic flux through the ring. Let us assume for example that this flux is downward through the ring. By Lenz s law this change in netic flux through the ring will give rise to a uced B-field that will be upward through the ring. Therefore it is as though this uced B-field were produced by a bar net with its south pole at the bottom of the ring and the north pole at the top of the ring. The south pole of this B repels the south pole of the B due to the current in the coil. This repulsive force makes the ring fly off. CQ9. Anytime a conductor is moved in a B-field there is an emf uced in the conductor. The wire is pulled through the B-field of the Earth and this generates an emf in the wire. f this wire were then connected to an external circuit the wire would act like a battery pushing charge through the external circuit. roblems. Let us refer to the bottom surface of the box as ide and the lateral surfaces (the sides) as ides 3 4 and 5. (t will not matter which is which is 3 and so on.) The netic flux through any surface is defined by Φ BAcosθ in which θ is the angle between B and the normal to the surface you re talking about. Looking at the figure we see that everywhere on ides 3 4 and 5 the normal points horizontally. The B-field is vertically upward. Therefore the angle θ is 90 and the flux through these sides is zero: Φ Φ Φ Φ BAcos 90 0. ( ) ( ) ( ) ( ) 3 4 5 For ide we he: ( Φ ) BAcosθ. For ide the normal is either vertically upward (giving θ 0 ) or vertically downward (giving θ 80 ). ince m interested in the nitude of the flux ll oid the minus sign by taking the normal to point upward. Then: ( ) cos 0 Φ BA BA The area A is the area of ide. From the figure this is LW. o we he: 4 ( Φ ) B( LW) ( )( )( ) 0.050 T 0.35 m 0.0 m 9.75 0 Wb 9. The problem should say what is the nitude of the erage uced emf? The absolute value of the erage uced emf is given by Faraday s law:. ( BAcosθ )
hysics 40 Homework olutions - Walker Chapter 3 The angle θ is the angle between B and the normal to the cross-sectional area of the loop. This angle is either 0 or 80. will take it to be 0 to oid a minus sign since m interested in the absolute value of the uced emf. The area of the loop is not changing so A B and Faraday s law becomes: B A ( ) π ( ) 0.5 T 53 0.5 m 7.8 t 0. s in which ve used A π r for the area of the circular loop.. since the number of turns is. (t s a single-loop coil the problem says.) The trick here is to realize what represents on a graph of Φ vs t: it represents the slope of the graph. (a.) At t 0.050 s we are on a segment of the graph that has slope 0 Wb 0 Wb 00. 0. s 0 s ( used the two endpoints of this segment to calculate the slope since know exactly where those endpoints are.) Therefore at t 0.050 s ( ) 00 ε. (b.) At t 0.5 s we re on a segment of the graph that has a slope of zero. (This segment is horizontal.) ε. (The flux is not changing during this segment therefore there is no uced emf.) Therefore ( ) 0 (c.) At t 0.50 s we re on a segment of the graph that has slope 5 Wb 0 Wb 8 0.6 s 0. s Therefore ( ) ( ) ε 38 + 38 8. The B-field due to the net is downward and getting larger as the net falls toward the ring. Therefore by Lenz s law this will cause an uced B-field B that is upward in the interior of the ring to partially cancel some of the increase in the downward flux due to the net. By the H for the sense in which B circulates around a current-carrying wire if you iine grasping the ring anywhere in such a way that your
hysics 40 Homework olutions - Walker Chapter 3 fingers curl around and point upward in the interior of the ring your thumb will he to point in the counterclockwise sense. Therefore must flow counterclockwise as viewed from above. 7. For ing A: By the H for the sense in which B circulates around a current-carrying wire the B-field due to the wire is out of the page through ing A. Furthermore if the current in the wire is increasing with time this B-field is getting larger. Therefore by Lenz s law B will point into the page through ing A. By the H for the sense in which B circulates around a current-carrying wire this means that will he to flow clockwise around ing A. will come back to ing B in just a moment. But first let s take care of ing C. For ing C: The B-field due to the wire is into the page and increasing. Therefore by Lenz s law B will be out of the page through ing C. This means that will he to flow counterclockwise around ing C. For ing B: Think of ing B as consisting of two halves one above the wire and the other below the wire. For the half of ing B that s above the wire the flux due to the wire is out of the page. For the half of ing B that s below the wire the flux due to the wire is into the page. Furthermore because the area of the top half of ing B is the same as the area of the bottom half Φ through the top half is equal in nitude to Φ through the bottom half. This means that the total flux through ing B is zero. And it s zero at all times! As the current in the wire increases the B-field through the top half gets bigger so Φ through the top half gets bigger. But Φ through the bottom half gets bigger by the same amount so the total flux through the ring remains zero. By Faraday s law ( ε ) if Φ is not changing with time there will be no uced emf in ing B. Therefore no uced current flows in ing B. 30. (a.) ε BLv ( 0.5 A)(.5 Ω) ( 0.750 T)( 0.45 m) v 4.6 m/s BL (b.) f the bar were moving to the left the polarity of the uced emf would be swapped. With the bar moving to the right the bottom of the bar is at higher potential than the top of the bar. f the bar moved to the left the top of the bar would be at higher potential than the bottom. But it would still be true that ε BLv. Therefore we would still get 3
hysics 40 Homework olutions - Walker Chapter 3 BLv and this would still lead to v BL as before. Therefore there would be no change in the speed required. 3. (a.) Once some current begins to flow the rod becomes a current-carrying wire in a B-field. Therefore there s a netic force exerted on the rod given by F LBsinθ in which θ means the angle between and B. The current flows downward through the rod; B is out of the page. Therefore θ 90 and we he F LB By the H for the direction of this force points to the left if the rod is moving to the right. Therefore opposes the motion and would cause the rod to slow down were it not for some other force an externally force continuing to pull the bar to the right. Let us call this externally force. And now we come to the bit about staying constant. f is to stay constant the bar must be pulled with constant speed. (ee the expression for in roblem 30 if this isn t immediately clear.) Well if the rod is to be pulled with constant speed the force must equal in nitude! Therefore F LB ( )( )( ) F 0.5 A 0.45 m 0.750 T 0.04 to the right. (b.) The rate of energy dissipation in the resistor is the power dissipated in the resistor. This is given by ε o: BLv B L v ( BLv) ( 0.750 T) ( 0.45 m) ( 4.6 m/s) 0.0 W.5 Ω 4
hysics 40 Homework olutions - Walker Chapter 3 (c.) f the rod is not to slow down then the externally force must deliver an amount of power equal to the power being dissipated in the resistor. o the mechanical power delivered to the rod must be 0.0 W. 39. Faraday s law written in terms of the uctance L says L t 55 0 A 45.0 0 H.4 6.5 0 s ( ) ( ) ε 4. Assuming the solenoid to be an ideal solenoid (and they should he said this) the uctance L is given by µ 0 L A ( ) L 7 T m 640 4 π 0 ( 0.043 m) 0.0 H A 0.5 m π in which ve used the fact that the area of a circle is π r. 6. We re told that 5 and 750. We want 4800 and 0.0 A. What do need to be? Well 750 ( 0.0 A) 0.36 A 5 And 5 ( 4800 ) 60 750 5