Two concentric circular loops of radius 5 cm and 15 cm are placed co-axially. Find e.m.f induced in the inner loop, if the current through the outer loop is changed at a rate of 10 A/ms. Assume the magnetic field on the inner loop to be uniform.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( -3.29\times10^{-4}\,\mathrm{V} \)
The current in the outer loop sets up a field at its centre; the inner loop is small, so this field is taken as uniform over the inner loop, as the question says. A changing current changes the flux through the inner loop, and Faraday's law gives the emf, with the negative sign from Lenz's law.
\( \displaystyle B = \frac{\mu_0 I}{2R} \)
\( \displaystyle \Phi = B\,\pi r^{2} = \frac{\mu_0 \pi r^{2}}{2R}\,I \)
\( \displaystyle \varepsilon = -\frac{d\Phi}{dt} = -\frac{\mu_0 \pi r^{2}}{2R}\,\frac{dI}{dt} \)
\( R = 15\,\mathrm{cm} = 0.15\,\mathrm{m} \)
\( r = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( \displaystyle \frac{dI}{dt} = \frac{10\,\mathrm{A}}{10^{-3}\,\mathrm{s}} = 10^{4}\,\mathrm{A\,s^{-1}} \)
\( \displaystyle \varepsilon = -\frac{4\pi\times10^{-7} \times \pi \times (0.05)^{2}}{2 \times 0.15} \times 10^{4} \)
\( \displaystyle \varepsilon = -\frac{4\pi^{2}\times10^{-7} \times 2.5\times10^{-3}}{0.3} \times 10^{4} \approx -3.29\times10^{-4}\,\mathrm{V} \)