A square loop of wire with side length 10 cm is placed at an angle of 60° with a magnetic field that changes uniformly from 0.2 T to zero in 0.1 s. If the loop resistance is 0.1 Ω, the induced current in the loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.2 A
By Faraday’s law, the emf equals the rate of change of magnetic flux through the loop. The flux uses the component of B along the normal to the loop: the loop’s plane makes 60° with the field, so the normal makes 90° − 60° = 30° with it.
\( A = (10\,\mathrm{cm})^{2} = (0.10\,\mathrm{m})^{2} = 1.0\times10^{-2}\,\mathrm{m^{2}} \)
\( \theta = 30^\circ \text{ (between B and the normal)} \)
\( \displaystyle \left|\frac{\Delta B}{\Delta t}\right| = \frac{0.2\,\mathrm{T} - 0}{0.1\,\mathrm{s}} = 2\,\mathrm{T\,s^{-1}} \)
\( \phi = BA\cos\theta \)
\( \displaystyle \varepsilon = A\cos\theta \left|\frac{\Delta B}{\Delta t}\right| = 1.0\times10^{-2}\,\mathrm{m^{2}} \times 0.866 \times 2\,\mathrm{T\,s^{-1}} \)
\( \varepsilon = 1.73\times10^{-2}\,\mathrm{V} \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{1.73\times10^{-2}\,\mathrm{V}}{0.1\,\Omega} \)
\( I = 0.173\,\mathrm{A} \approx 0.2\,\mathrm{A} \)
Note: In the Final Answer Key, NTA accepted both Option (2) and Option (3). Option (3), printed by NTA as 0.1 A, is what one gets by taking 60° as the angle between the field and the normal to the loop (\( \phi = BA\cos 60^\circ \), \( \varepsilon = 0.01\,\mathrm{V} \), I = 0.1 A). For the benefit of students and to ensure a single answer, Option (3) has been changed from 0.1 A to 0.4 A.