Homework 10 Solutions

# 17 equation 29 10 gives the field at the center of

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Unformatted text preview: (counterclockwise). 16. (a) Since the flux arises from a dot product of vectors, the result of one sign for B1 and B2 and of the opposite sign for B3 (we choose the minus sign for the flux from B1 and B2, and therefore a plus sign for the flux from B3). The induced emf is dΦB ⎛dB1 dB2 dB3⎞ ε = −Σ dt = A ⎜ dt + dt − dt ⎟ ⎠ ⎝ =(0.10 m)(0.20 m)(2.0 × 10−6 T/s + 1.0 ×10−6 T/s −5.0×10−6 T/s) = −4.0×10−8 V. The minus sign means that the effect is dominated by the changes in B3. Its magnitude (using Ohm’s law) is |ε| /R = 8.0 μA. (b) Consideration of Lenz’s law leads to the conclusion that the induced current is therefore counterclockwise. 17. Equation 29-10 gives the field at the center of the large loop with R = 1.00 m and current i(t). This is approximately the field throughout the area (A = 2.00 × 10–4 m2) enclosed by the small loop. Thus, with B = μ0i/2R and i(t) = i0 + kt, where i0 = 200 A and k = (–200 A – 200 A)/1.00 s = – 400 A/s, we find (a) B (t = 0) = μ 0i0 2R ( 4π×10 = ( 4π×10 (b) B (t = 0.500s) = −7 H/m ) ( 200A ) 2 (1.00m ) −7 = 1.26 ×10−4 T, H/m ) ⎡ 200A − ( 400A/s ) ( 0.500s ) ⎤ ⎣ ⎦ 2 (1.00m ) = 0, and 1171 be in radians (here, ω = 2πf is the angular velocity of the coil in radians per second, and f = 1000 rev/min ≈ 16.7 rev/s is the frequency). Since the area of the rectangular coil is A = (0.500 m) × (0.300 m) = 0.150 m2, Faraday’s law leads to ε = −N b g bg d BA cosθ d cos 2 πft = − NBA = NBA2 πf sin 2 πft dt dt bg which means it has a voltage amplitude of ε max = 2π fNAB = 2π (16.7 rev s ) (100 turns ) ( 0.15 m 2 ) ( 3.5T ) = 5.50 ×103 V . 20. We note that 1 gauss = 10–4 T. The amount of charge is N 2 NBA cos 20° [ BA cos 20° − (− BA cos 20°)] = R R −4 2 2(1000)(0.590 × 10 T)π(0.100 m) (cos 20°) = = 1.55 ×10−5 C . 85.0 Ω + 140 Ω q (t ) = Note that the axis of the coil is at 20°, not 70°, from the magnetic field of the Earth. 21. (a) The frequency is f= ω (40 rev/s)(2π rad/rev) = = 40 Hz . 2π 2π (b) First, we define angle relative to the plane of Fig. 30-44, such that the semicircular wire is in the θ = 0 position and a quarter of a period (of revolution) later it will be in the θ = π/2 position (where its midpoint will reach a distance of a above the plane of the figure). At the moment it is in the θ = π/2 position, the area enclosed by the “circuit” will appear to us (as we look down at the figure) to that of a simple rectangle (call this area A0, which is the area it will again appear to enclose when the wire is in the θ = 3π/2 position). Since the area of the semicircle is πa2/2, then the area (as it appears to us) enclosed by the circuit, as a function of our angle θ, is A = A0 + πa 2 cosθ 2 where (since θ is increasing at a steady rate) the angle depends linearly on time, which we can write either as θ = ωt or θ = 2πft if we take t = 0 to be a moment when the arc is in the θ = 0 position. Since B is uniform (in space) and consta...
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