JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A square loop of side \(20\, {cm}\) and resistance \(1\, \Omega\) is moved towards right with a constant speed \({v}_{0}\). The right arm of the loop is in a uniform magnetic field of \(5\, {T}\). The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value \(4\, \Omega\). What should be the value of \(v_{0}\) so that a steady current of \(2\, {mA}\) flows in the loop ?

- A \(1\, {m} / {s}\)
- B \(1 \,{cm} / {s}\)
- C \(10^{2}\, {m} / {s}\)
- D \(10^{-2} \,{cm} / {s}\)
Answer & Solution
Correct Answer
(B) \(1 \,{cm} / {s}\)
Step-by-step Solution
Detailed explanation
Equivalent circuit \(i=\frac{{VoB} \ell}{4+1} \Rightarrow {V}_{0}=\frac{5(2\, {mA})}{5 \times .2}=10^{-2} \,{m} / {s}=1\, {cm} / {s}\)
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