Q 12-06-052JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
A rectangular loop of length $2.5\ \text{m}$ and width $2\ \text{m}$ is placed at $60^\circ$ to a magnetic field of $4\ \text{T}$. The loop is removed from the field in $10\ \text{s}$. The average emf induced in the loop during this time is:
Answer: (C) $+1\ \text{V}$
Initial flux (taking the $60^\circ$ as the angle between the field and the normal of the loop):
$$\Phi_i = BA\cos60^\circ = 4\times(2.5\times2)\times\frac12 = 10\ \text{Wb},\qquad \Phi_f = 0$$
$$\varepsilon_{avg} = -\frac{\Delta\Phi}{\Delta t} = -\frac{0-10}{10} = +1\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics