9. (1.5 points) The figure shows a cross section of three parallel wires each ca

Question

9. (1.5 points) The figure shows a cross section of three parallel wires each carrying a current of 20 The currents in wire A and B are out of the paper, while that in wire C is into the paper. If the distance R = 5.0 mm, what is the magnitude of the force on a 0.5-m length of wire A? 23 mN 8mN 32 mN 5 mN 55mN 3.(1.5 points) A square loop (length) along one side = 15 cm) rotates is a constant magnetic field which has a magnitude of 2.0 T. At an instant when the angle between the field and the normal to the plane of the loop is equal to 20 degree and increasing at the rate of 10 degree/s, what is the magnitude of the induced emf in the loop? (360 degree = 2pie rad) 10 mV 2.7 mV 4.8 mV 14 mV 2.2 mV

Explanation / Answer

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net Force due to two same direction currents is zero

so Force on wire at A is only due to current of C

so magnetic force F = ILB

F = 20 *0.5 * B

B = uoi/2pIR

B = 4pie-7 * 20/(2* 3.14* 5e-3)

B = 8 e- 4 T

so F = 20 * 0.5* 8 e-4

F = 8 mT

option B it is
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induced emf e =- NABW cos theta/dt

emf e = 1* 0.15*0.15*2   * sin 20 * ( 10* 2pi/360)

emf =e = 2.67 mV

option b it is
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