JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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
- A At \(t=0.5\) direction of \(I_{R}\) reverses and \(V_{R}\) is zero
- B Direction of \(I_R\) remains unchanged and \(\mathrm { V } _ { \mathrm { R } }\) is zero at \(t=0.25\)
- C Direction of \(I_R\) remains unchanged and \(V_R \) is maximum at \(t=0.5\)
- D At \(t=0.25\) direction of \(I_{R}\) reverses and \(V_{R}\) is maximum
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…
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