A coil of 500 turns is threaded by a magnetic flux of \( 7\times10^{-5}\,\mathrm{Wb} \). If the flux changes to \( 3\times10^{-5}\,\mathrm{Wb} \) in 0.01 s, what average induced emf is developed in the coil? If resistance offered by the coil is \( 10\,\Omega \), then the flow of average induced current will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2 V, 0.2 A
By Faraday’s law the average induced emf in a coil of N turns is \( \varepsilon = N\,\frac{|\Delta\Phi|}{\Delta t} \). The current then follows from Ohm’s law.
\( |\Delta\Phi| = (7-3)\times10^{-5} = 4\times10^{-5}\,\mathrm{Wb} \)
\( \displaystyle \varepsilon = 500\times\frac{4\times10^{-5}}{0.01} = 2\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{2\,\mathrm{V}}{10\,\Omega} = 0.2\,\mathrm{A} \)