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JEE MAIN_2026
05-04-26_S2
Question
A circular loop of radius 20 cm and resistance $2 \Omega$ is placed in a time varying magnetic field $\vec{B}=\left(2 t^2+2 t+3\right) T$. At $\mathrm{t}=0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t=3 \mathrm{~s}$ is $\frac{\alpha}{50} \mathrm{~A}$. The value of $\alpha$ is $\_\_\_\_$ . (Take $\pi=22 / 7$ )
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Solution
$\begin{aligned} & \phi=\mathrm{B} \cdot \pi \mathrm{r}^2 \\ & \varepsilon=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\pi \mathrm{r}^2 \frac{\mathrm{~dB}}{\mathrm{dt}} \\ & \mathrm{i}=\frac{\varepsilon}{\mathrm{R}}=\frac{\pi \mathrm{r}^2}{\mathrm{R}} \cdot\left(\frac{\mathrm{dB}}{\mathrm{dt}}\right)=\frac{\pi}{2}\left(\frac{1}{5}\right)^2(4 \mathrm{t}+2) \\ & \text { for } \mathrm{t}=3 \\ & i=\frac{11}{7} \times \frac{1}{25} \times 14 \\ & i=\frac{44}{50}\end{aligned}$
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