MHT CET · Physics · Electromagnetic Induction
A rectangular loop \(\mathrm{PQMN}\) with movable arm \(\mathrm{PQ}\) of length \(12 \mathrm{~cm}\) and resistance \(2 \Omega\) is placed in a uniform magnetic field of \(0.1 \mathrm{~T}\) acting perpendicular to the plane of the loop as shown in figure. The resistance of the arms MN, NP and MQ are negligible. The current induced in the loop when arm \(\mathrm{PQ}\) is moved with velocity \(20 \mathrm{~ms}^{-1}\) is
- A \(0.12 \mathrm{~A}\)
- B \(0.06 \mathrm{~A}\)
- C \(0.24 \mathrm{~A}\)
- D \(0.18 \mathrm{~A}\)
Answer & Solution
Correct Answer
(A) \(0.12 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}=0.1 \mathrm{~T}, \ell=0.12 \mathrm{~m}, \mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{R}=2 \Omega\)
emf induced \(\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.1 \times 0.12 \times 20=0.24 \mathrm{~V}\)
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{0.24}{2}=0.12 \mathrm{~A}\)
emf induced \(\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.1 \times 0.12 \times 20=0.24 \mathrm{~V}\)
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{0.24}{2}=0.12 \mathrm{~A}\)
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