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JEE Mains · Physics · STD 12 -6. Electromagnetic induction
An infinitely long straight wire carrying current \(I,\) one side opened rectangular loop and a conductor \(C\) with a sliding connector are located in the same plane, as shown in the figure. The connector has length \(l\) and resistance \(R\). It slides to the right with a velocity \(V.\) The resistance of the conductor and the self inductance of the loop are negligible. The induced current in the loop, as a function of separation \(r,\) between the connector and the straight wire is
- A \(\frac{\mu_{0}}{\pi} \frac{ Iv l}{ Rr }\)
- B \(\frac{\mu_{0}}{2 \pi} \frac{ Ivl }{ Rr }\)
- C \(\frac{2 \mu_{0}}{\pi} \frac{ Ivl }{ Rr }\)
- D \(\frac{\mu_{0}}{4 \pi} \frac{ Iv l}{ Rr }\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_{0}}{2 \pi} \frac{ Ivl }{ Rr }\)
Step-by-step Solution
Detailed explanation
\(B =\frac{\mu_{0} i }{2 \pi r }\) \(\phi=\frac{\mu_{0} i }{2 \pi r } \ell dr\) \(\Rightarrow \frac{ d \phi}{ dt }=\frac{\mu_{0} i \ell}{2 \pi r } \cdot \frac{ dr }{ dt }\) \(\Rightarrow e =\frac{\mu_{0}}{2 \pi} \cdot \frac{ iv \ell}{ r }\)…
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