CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Maximum value of induced emf is 0.422 V
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Maximum value of induced emf is 0.122 V
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

Field due to the toroid is,

B=μoNI2πr=500μoI2πr

Flux linked with the coil is,

ϕB=BdA=500μ0Imaxsinωt2πb+RRadrr

ϕB=500μ0Imax2πasinωt ln (b+RR)
E=NdϕBdt=20(500μ0Imax2π)ω aln (b+RR)cosωt
E=1042π(4π×107)(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.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Faraday’s Law of Induction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon