A conducting circular loop is placed in a uniform magnetic field, \( B = 0.025\,\mathrm{T} \) with its plane perpendicular to the magnetic field. The radius of the loop is made to shrink at a constant rate of \( 2\,\mathrm{mm\,s^{-1}} \). The induced emf when the radius is 2 cm, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\pi\,\mu\mathrm{V} \)
By Faraday's law the induced emf equals the rate of change of magnetic flux. The loop is perpendicular to the field, so \( \Phi = B A = B\pi r^{2} \), and only the area changes.
\( \displaystyle |\varepsilon| = \frac{d\Phi}{dt} = B\,\frac{d(\pi r^{2})}{dt} = 2\pi r B\,\frac{dr}{dt} \)
\( r = 2\,\mathrm{cm} = 2\times10^{-2}\,\mathrm{m} \)
\( dr/dt = 2\,\mathrm{mm\,s^{-1}} = 2\times10^{-3}\,\mathrm{m\,s^{-1}} \)
\( B = 0.025\,\mathrm{T} \)
\( \displaystyle |\varepsilon| = 2\pi \times 2\times10^{-2} \times 0.025 \times 2\times10^{-3} = 2\pi\times10^{-6}\,\mathrm{V} = 2\pi\,\mu\mathrm{V} \)