A square loop of wire of side \( 2 \) m is placed inside a large circular loop of wire of radius \( 8 \) m. The loops are co-planar and their centers coincide. If the current \( I \) passes through the circular loop, the mutual inductance of the system is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3.14\times10^{-7} \) H
The mutual inductance is the flux through one loop per unit current in the other: \( M = \Phi_2/I_1 \). Let the current I flow in the large circular loop.
The square (side 2 m) is small compared with the circle (radius 8 m) and sits at its centre, so the field of the circular loop over the square is taken as uniform and equal to its value at the centre.
\( \displaystyle B = \frac{\mu_0I}{2R} \)
\( \displaystyle \Phi = B\,a^{2} = \frac{\mu_0I}{2R}\,a^{2} \)
\( \displaystyle M = \frac{\Phi}{I} = \frac{\mu_0a^{2}}{2R} = \frac{(4\pi\times10^{-7})(2)^{2}}{2\times8} \)
\( \displaystyle M = \frac{16\pi\times10^{-7}}{16} = \pi\times10^{-7} \approx 3.14\times10^{-7}\,\mathrm{H} \)