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 .....\(\mu V\)

Pick an option to check your answer.
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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