/R = If we take the surface normal to be +ẑ, then the right hand rule gives positive flowing current to be in the +ˆφ direction.

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Moris Cummings
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1 Problem 6.2 The loop in Fig. P6.2 is in the x y plane and B = ẑb 0 sinωt with B 0 positive. What is the direction of I (ˆφ or ˆφ) at: (a) t = 0 (b) ωt = π/4 (c) ωt = π/2 V emf y x I Figure P6.2: Loop of Problem 6.2. Solution: I = V emf /. Since the single-turn loop is not moving or changing shape with time, Vemf m = 0 V and V emf = Vemf tr. Therefore, from Eq. (6.8), I = Vemf tr 1 B / = S t ds. If we take the surface normal to be +ẑ, then the right hand rule gives positive flowing current to be in the +ˆφ direction. I = A t B 0 sinωt = AB 0ω cosωt where A is the area of the loop. (a) A, ω and are positive quantities. At t = 0, cosωt = 1 so I < 0 and the current is flowing in the ˆφ direction (so as to produce an induced magnetic field that opposes B). (b) At ωt = π/4, cosωt = 2/2 so I < 0 and the current is still flowing in the ˆφ direction. (c) At ωt = π/2, cosωt = 0 so I = 0. There is no current flowing in either direction. (A),

2 Problem 6.4 A stationary conducting loop with an internal resistance of 0.5 Ω is placed in a time-varying magnetic field. When the loop is closed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2-Ω resistor is connected across its open ends? Solution: V emf is independent of the resistance which is in the loop. Therefore, when the loop is intact and the internal resistance is only 0.5 Ω, V emf = 5 A 0.5 Ω = 2.5 V. When the small gap is created, the total resistance in the loop is infinite and the current flow is ero. With a 2-Ω resistor in the gap, I = V emf /(2 Ω+0.5 Ω) = 2.5 V/2.5 Ω = 1 (A).

3 Problem 6.6 The square loop shown in Fig. P6.6 is coplanar with a long, straight wire carrying a current I(t) = 5cos(2π 10 4 t) (A). (a) Determine the emf induced across a small gap created in the loop. (b) Determine the direction and magnitude of the current that would flow through a 4-Ω resistor connected across the gap. The loop has an internal resistance of 1 Ω. I(t) 5 cm 10 cm 10 cm y x Figure P6.6: Loop coplanar with long wire (Problem 6.6). Solution: (a) The magnetic field due to the wire is B = ˆφ µ 0I 2πr = ˆx µ 0I 2πy, where in the plane of the loop, ˆφ = ˆx and r = y. The flux passing through the loop is 15 cm ( Φ = B ds = ˆx µ ) 0I [ ˆx10 (cm)] dy S 5 cm 2πy = µ 0I 10 1 ln 15 2π 5 = 4π cos(2π 10 4 t) π = cos(2π 10 4 t) (Wb).

4 V emf = dφ = 1.1 2π 10 4 sin(2π 10 4 t) 10 7 dt = sin(2π 10 4 t) (V). (b) I ind = V emf = sin(2π 10 4 t) = 1.38sin(2π 10 4 t) (ma). At t = 0, B is a maximum, it points in ˆx-direction, and since it varies as cos(2π 10 4 t), it is decreasing. Hence, the induced current has to be CCW when looking down on the loop, as shown in the figure.

5 Problem 6.7 The rectangular conducting loop shown in Fig. P6.7 rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B = ŷ50 (mt). Determine the current induced in the loop if its internal resistance is 0.5 Ω. 2 cm 3 cm B B y x φ(t) ω Figure P6.7: otating loop in a magnetic field (Problem 6.7). Solution: Φ = B ds = ŷ ŷ( )cosφ(t) = cosφ(t), S 2π φ(t) = ωt = t = 200πt (rad/s), 60 Φ = cos(200πt) (Wb), V emf = dφ = π sin(200πt) = sin(200πt) (V), dt I ind = V emf 0.5 = 37.7sin(200πt) (ma). The direction of the current is CW (if looking at it along ˆx-direction) when the loop is in the first quadrant (0 φ π/2). The current reverses direction in the second quadrant, and reverses again every quadrant.

6 ω Problem 6.10 A 50-cm-long metal rod rotates about the -axis at 90 revolutions per minute, with end 1 fixed at the origin as shown in Fig. P6.10. Determine the induced emf V 12 if B = ẑ T. B 1 y x 2 Figure P6.10: otating rod of Problem Solution: Since B is constant, V emf = Vemf m. The velocity u for any point on the bar is given by u = ˆφrω, where From Eq. (6.24), ω = 2π rad/cycle (90 cycles/min) (60 s/min) = 3π rad/s. 1 0 V 12 = Vemf m = (u B) dl = (ˆφ3πr ẑ ) ˆr dr 2 r=0.5 0 = 6π 10 4 r dr r=0.5 = 3π 10 4 r = 3π = 236 (µv).

7 Problem 6.11 The loop shown in P6.11 moves away from a wire carrying a current I 1 = 10 A at a constant velocity u = ŷ7.5 (m/s). If = 10 Ω and the direction of I 2 is as defined in the figure, find I 2 as a function of y 0, the distance between the wire and the loop. Ignore the internal resistance of the loop. 10 cm I 1 = 10 A u 20 cm I 2 u y 0 Figure P6.11: Moving loop of Problem Solution: Assume that the wire carrying current I 1 is in the same plane as the loop. The two identical resistors are in series, so I 2 = V emf /2, where the induced voltage is due to motion of the loop and is given by Eq. (6.26): V emf = V m (u B) dl. emf = C The magnetic field B is created by the wire carrying I 1. Choosing ẑ to coincide with the direction of I 1, Eq. (5.30) gives the external magnetic field of a long wire to be B = ˆφ µ 0I 1 2πr. For positive values of y 0 in the y- plane, ŷ = ˆr, so u B = ŷ u B = ˆr u ˆφ µ 0I 1 2πr = ẑ µ 0I 1 u 2πr. Integrating around the four sides of the loop with dl = ẑ d and the limits of integration chosen in accordance with the assumed direction of I 2, and recogniing

8 that only the two sides without the resistors contribute to Vemf m, we have 0 ) 0.2 ( Vemf m = ẑ µ ) 0I 1 u 0 2πr r=y 0 (ẑ d)+ = 4π π ( ) 1 = y 0 y ( 1 y 0 (V), ( ẑ µ 0I 1 u 2πr 1 y ) (ẑ d) r=y and therefore I 2 = V emf m ( ) 1 2 = y 0 y (na).

I.~ I ./ TT. Figure P6.1: Loops of Problem 6.1.

I.~ I ./ TT. Figure P6.1: Loops of Problem 6.1. CHAPTER 6 249 Chapter 6 Sections 6-1 to 6-6: Faraday's Law and its Applications Problem 6.1 The switch in the bottom loop of Fig. 6-17 (p6.1) is closed at t = 0 and then opened at a later time tl. What