JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 10)
A conducting circular loop of radius $$\frac{10}{\sqrt\pi}$$ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is :
emf = 10 mV
emf = 5 mV
emf = 100 mV
emf = 1 mV
Explanation
$\begin{aligned} & \mathrm{EMF}=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{BA}-0}{\mathrm{t}} \\\\ & \mathrm{A}=\pi \mathrm{r}^2=\pi\left(\frac{0.1^2}{\pi}\right)=0.01 \\\\ & \mathrm{~B}=0.5 \\\\ & \mathrm{EMF}=\frac{(0.5)(0.01)}{0.5}=0.01 \mathrm{~V}=10~ \mathrm{mV}\end{aligned}$
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