JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2023
- Hard
- MCQ
A square loop of area \(25\,cm ^2\) has a resistance of \(10\,\Omega\). The loop is placed in uniform magnetic field of magnitude \(40.0 T\). The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in \(1.0 sec\), will be \(..........\times 10^{-3}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(1.0\)
Step-by-step Solution
Detailed explanation
\(\ell=50\,cm\)
\(t =1\,sec\)
\(\therefore V =\frac{0.05}{1}=0.05\,m / s\)
\(i =\frac{40 \times 0.05 \times 0.05}{10}=0.01\,A\)
\(F = B { i } l=40 \times 0.01 \times 0.05\)
\(F =0.02\,N\)
\(\therefore \quad W =0.02 \times l=0.02 \times .05\)
\(\therefore \quad W =1 \times 10^{-3}\,J\)
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