Question

# A toroid having a rectangular cross-section (a=2.00 cm and b=3.00 cm) and inner radius r=4.00 cm consists of 500 turns of wire that carries a current I=Imaxsinωt with Imax=50.0 A and a frequency f=ω2π=60.0 Hz. A coil that consists of 20 turns of wire links with the toroid as shown in figure. Then which of the following is/are correct?

A
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B
Maximum value of induced emf is 0.422 V
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C
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D
Maximum value of induced emf is 0.122 V
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Solution

## The correct option is C Field due to the toroid is, B=μoNI2πr=500μoI2πr Flux linked with the coil is, ϕB=∫BdA=500μ0Imaxsinωt2π∫b+RRadrr ϕB=500μ0Imax2πasinωt ln (b+RR) E=NdϕBdt=20(500μ0Imax2π)ω aln (b+RR)cosωt E=1042π(4π×10−7)(50)(377)(0.020)ln(3+44)cosωt E=(0.422)cosωt For maximum value of induced emf cosωt=1 E=0.422 V <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B), & (C) is the correct answer.

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