MHT CET · Physics · Electromagnetic Induction
A coil of 'n' turns and resistance 'R' \(\Omega\) is connected in series with a resistance \(\frac{\mathrm{R}}{2}\) The combination is moved for time 't' second through magnetic flux \(\phi_{1}\) to \(\phi_{2}\). The induced current in the circuit is
- A \(\frac{2 n\left(\phi_{1}-\phi_{2}\right)}{3 R t}\)
- B \(\frac{n\left(\phi_{1}-\phi_{2}\right)}{3 R t}\)
- C \(\frac{n\left(\phi_{1}-\phi_{2}\right)}{R t}\)
- D \(\frac{2 n\left(\phi_{1}-\phi_{2}\right)}{R t}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 n\left(\phi_{1}-\phi_{2}\right)}{3 R t}\)
Step-by-step Solution
Detailed explanation
Total resistance \(=\mathrm{R}+\frac{\mathrm{R}}{2}=\frac{3 \mathrm{R}}{2}\)
\(\mathrm{i}_{\text {induced }}=\frac{\mathrm{e}_{\text {induced }}}{\frac{3 \mathrm{R}}{2}}=\frac{\frac{\mathrm{d} \phi}{\mathrm{dt}}}{\frac{3 \mathrm{R}}{2}}=\frac{2 \mathrm{n}}{3 \mathrm{Rt}} \frac{\left(\phi_{1}-\phi_{2}\right)}{}\)
\(\mathrm{i}_{\text {induced }}=\frac{\mathrm{e}_{\text {induced }}}{\frac{3 \mathrm{R}}{2}}=\frac{\frac{\mathrm{d} \phi}{\mathrm{dt}}}{\frac{3 \mathrm{R}}{2}}=\frac{2 \mathrm{n}}{3 \mathrm{Rt}} \frac{\left(\phi_{1}-\phi_{2}\right)}{}\)
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