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

A conducting ring of mass 2 kg and radius 0.5 m is placed on a smooth horizontal plane. The ring carries a current of i=4 A. A horizontal magnetic field B=10 T is switched on at time t=0 as shown in fig. The initial angular acceleration of the ring will be

166427.png

A
40π rad/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
20π rad/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
5π rad/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
15π rad/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 20π rad/s2
Let B=B^i
Magnetic moment μ=iA(^k)=iπr2(^k)

τ=μ×B=iπr2(^k)×B^i=iπr2B(^j).......................................(1)

But, τ=Iα where I=mr2 is the moment of inertia and α is the angular acceleration

τ=mr2α..............................(2)

Equating eqns (1) and (2), iπr2B=mr2α

α=iπBm=4×π×102=20πrad s2

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Moment of Inertia of Solid Bodies
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon