JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 25)
A moving coil galvanometer has 100 turns and each turn has an area of $2.0 \mathrm{~cm}^2$. The magnetic field produced by the magnet is $0.01 \mathrm{~T}$ and the deflection in the coil is 0.05 radian when a current of $10 \mathrm{~mA}$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad}$. The value of $x$ is _______ .
Answer
4
Explanation
$\begin{aligned} & \tau=\text { BINAsin } \phi \\\\ & \mathrm{C} \theta=\text { BINAsin } 90^{\circ} \\\\ & \mathrm{C}=\frac{\mathrm{BINA}}{\theta}=\frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05} \\\\ & =4 \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad} . \\\\ & \mathrm{x}=4\end{aligned}$
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