Q 12-06-059JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A coil is placed perpendicular to a magnetic field of $5000\ \text{T}$. When the field is changed to $3000\ \text{T}$ in $2\ \text{s}$, an induced emf of $22\ \text{V}$ is produced in the coil. If the diameter of the coil is $0.02\ \text{m}$, then the number of turns in the coil is:
Answer: (B) $70$
$A = \pi(0.01)^2 = \pi\times10^{-4}\ \text{m}^2$, $\dfrac{\Delta B}{\Delta t} = \dfrac{2000}{2} = 1000\ \text{T s}^{-1}$.
$$\varepsilon = NA\frac{\Delta B}{\Delta t} \Rightarrow 22 = N\times\pi\times10^{-4}\times1000 \Rightarrow N = \frac{22}{0.1\pi} = 70$$
Solution by Sreeraj P, M.Sc Physics