wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a region of uniform magnetic induction B=102 tesla, a circular loop of radius 30 cm and resistance π2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the loop. If the loop rotates at 200 rpm , the maximum amplitude of the alternating current induced in the loop is

A
4π2 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
30 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
6 mA
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
200 mA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 6 mA
Maximum Amplitude of AC i0=V0R=ωNBAR=(2πv)NB(πr2)R
i0=2π×20060×1×102×π×(0.3)2π2=6 mA

flag
Suggest Corrections
thumbs-up
13
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon