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Physics · Current Electricity

JEE Main 2024 — 1 February, Shift 2 — Question 53

A moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm22.0 \mathrm{~cm}^{2}. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x×10−5 N−m/rad\mathrm{x} \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad}. The value of x is _______\_\_\_\_\_\_\_ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

τ=\tau= BINAsin ϕ\phi Cθ=BINAsin⁡90∘\mathrm{C} \theta=\mathrm{BINA} \sin 90^{\circ}

C=BINAθ=0.01×10×10−3×100×2×10−40.05\mathrm{C}=\frac{\mathrm{BINA}}{\theta}=\frac{0.01 \times 10 \times 10^{-3} \times 100 \times 2 \times 10^{-4}}{0.05}

=4×10−5 N−m/rad=4 \times 10^{-5} \mathrm{~N}-\mathrm{m} / \mathrm{rad}. x=4\mathrm{x}=4

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 2
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments

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