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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2014
  • Medium
  • Numerical

A square frame of side \(10\, cm\) and a long straight wire carrying current \(1\, A\) are in the plate of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of \(10\, ms^{-1}\) (see figure). The \(e.m. f\) induced at the time the left arm of the frame is at \(x\, = 10\, cm\) from the wire is .....\(\mu V\)

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Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

In the given question,

Current flowing through the wire, \(I=1\, A\)

Speed of the frame, \(\mathrm{v}=10\, \mathrm{ms}^{-1}\)

Side of square loop, \(l=10\, \mathrm{cm}\)

Distance of square frame from current carrying

wires \(x=10\, \mathrm{cm}\)

We have to find, \(e.m.f\) inducede =?

According to Biot-Savart's law

\(B = \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{{Idl\sin \,\theta }}{{{x^2}}}\)

\( = \frac{{4\pi  \times {{10}^{ - 7}}}}{{4\pi }} \times \frac{{1 \times {{10}^{ - 1}}}}{{{{({{10}^{ - 1}})}^2}}}\)

\(=10^{-6}\)

Induced \(e.m.f.\) \(\mathrm{e}=\mathrm{Blv}\)

\(=10^{-6} \times 10^{-1} \times 10\)

\(=1 \,\mu v\)

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