JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2024
- Hard
- Numerical
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 \(\mathrm{x} \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad}\). The value of \(\mathrm{x}\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\tau=\text { BINAsin } \phi\)
\(\mathrm{C} \theta=\mathrm{BINA} \sin 90^{\circ}\)
\(\mathrm{C}=\frac{\mathrm{BINA}}{\theta}=\frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05}\)
\(=4 \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad} .\)
\(\mathrm{x}=4\)
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