A coil with an average diameter of 0.02 m is placed with its plane perpendicular to a magnetic field of 6000 T. The induced emf in the coil is 11 V, when the magnetic field is changed to 1000 T in 4 s. The number of turns in the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 28
By Faraday's law, the magnitude of the induced emf in a coil of N turns is \( \varepsilon = N\,\Delta\Phi/\Delta t \), with \( \Phi = BA \) since the plane is perpendicular to the field.
\( r = \frac{0.02}{2}\,\mathrm{m} = 0.01\,\mathrm{m} \)
\( \displaystyle A = \pi r^{2} = 3.14 \times (0.01\,\mathrm{m})^{2} = 3.14\times10^{-4}\,\mathrm{m^{2}} \)
\( \displaystyle \frac{\Delta B}{\Delta t} = \frac{6000\,\mathrm{T} - 1000\,\mathrm{T}}{4\,\mathrm{s}} = 1250\,\mathrm{T\,s^{-1}} \)
\( \displaystyle N = \frac{\varepsilon}{A\,(\Delta B/\Delta t)} \)
\( \displaystyle N = \frac{11}{3.14\times10^{-4} \times 1250} \approx 28 \)