JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2020
- Hard
- MCQ
A long solenoid of radius \(\mathrm{R}\) carries a time \((t)-\)dependent current \(\mathrm{I}(\mathrm{t})=\mathrm{I}_{0} \mathrm{t}(1-\mathrm{t})\). A ring of radius \(2 \mathrm{R}\) is placed coaxially near its middle. During the time interval \(0 \leq t \leq 1\), the induced current \(\left(\mathrm{I}_{\mathrm{R}}\right)\) and the induced \(\mathrm{EMF}\left(\mathrm{V}_{\mathrm{R}}\right)\) in the ring change as
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) At \(t=0.5\) direction of \(I_{R}\) reverses and \(V_{R}\) is zero
Step-by-step Solution
Detailed explanation
Magnetic flux ( \(\phi\) ) through ring is \(\phi=\pi(R)^{2} . B\)
\(\phi=\left(\pi \mathrm{R}^{2}\right)\left(\mu_{0} \mathrm{n} \mathrm{I}\right)=\left(\pi \mathrm{R}^{2} \mu_{0} \mathrm{n} \mathrm{I}_{0}\right)\left(\mathrm{t}-\mathrm{t}^{2}\right)\)
Induced e.m.f. of \(\mathrm{V}_{\mathrm{R}}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}\)
\(=\left(\pi \mathrm{R}^{2} \mu_{0} \mathrm{nI}_{0}\right)(2 \mathrm{t}-1)\)
and induced current \(\mathrm{I}_{\mathrm{R}}=\frac{\pi \mathrm{R}^{2} \mu_{0} \mathrm{nI}_{0}(2 \mathrm{t}-1)}{\mathrm{R}_{\mathrm{R}}}\)
\(\left(\mathrm{R}_{\mathrm{R}} \rightarrow \text { Resistance of Ring }\right)\)
Clearly \(\mathrm{V}_{\mathrm{R}}\) and \(\mathrm{I}_{\mathrm{R}}\) are zero at \(\mathrm{t}=\frac{1}{2}=0.5 \mathrm{sec}\)
and their sign also changes at \(t=0.5\) sec.
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