A circular super conducting loop of area A and inductance L is located in a uniform field of flux density B. Plane of the loop is initially parallel to B and current in the loop is zero. The loop is now turned slowly through angle $\frac{π}{2}$ about its diameter such that its plane now becomes perpendicular to B. Work done by external agent in the process is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

zero

$\frac{B A}{L}$

$\frac{B A}{2 L}$

$\frac{B^{2} A^{2}}{2 L}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Flux associated with a super conducting loop remains unaltered $ϕ_{B} = L I_{i n d u c e d}$
Work $= \frac{1}{2} L i^{2} = \frac{1}{2} \frac{ϕ^{2}}{L^{2}} L = \frac{B^{2} A^{2}}{2 L}$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE