A coil in the shape of a square having 200 turns, with an area of each turn \( 2.5 \times 10^{-3}\,\mathrm{m^2} \) and a total resistance of \( 200\,\Omega \), is placed in a uniform magnetic field of 0.5 T, such that the plane of the coil is perpendicular to the magnetic field. The loop is pulled out of the field slowly and uniformly in 1.0 s as shown in the figure. The magnitude of induced current in the coil is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.25 mA
By Faraday's law, the emf induced in a coil equals the rate of change of the magnetic flux linked with it. The field is perpendicular to the plane of the coil, so the flux linked at the start is \( NBA \); after the coil leaves the field it is zero.
\( \displaystyle \Delta\Phi = N B A = 200 \times 0.5\,\mathrm{T} \times 2.5\times10^{-3}\,\mathrm{m^2} = 0.25\,\mathrm{Wb} \)
The coil is pulled out uniformly, so the emf is the change in flux divided by the time taken.
\( \displaystyle \varepsilon = \frac{\Delta\Phi}{\Delta t} = \frac{0.25\,\mathrm{Wb}}{1.0\,\mathrm{s}} = 0.25\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{0.25\,\mathrm{V}}{200\,\Omega} = 1.25\times10^{-3}\,\mathrm{A} = 1.25\,\mathrm{mA} \)