Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

194 of 276

Question 194 of 276

A moving coil galvanometer has 100100 turns and each turn has an area of 2.0cm2.2.0{cm}^{2}. The magnetic field produced by the magnet is 0.01T0.01T and the deflection in the coil is 0.050.05 radian when a current of 10mA10mA is passed through it. The torsional constant of the suspension wire is x×10−5N−m/rad.x\times {10}^{-5}N-m/rad. The value of xx is ______ .

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Answer: 4

Torque produced in a moving coil galvanometer is, τ=BINAsinϕ\tau =BINAsin\phi.
Let torsional constant of the suspension wire be CC, then Cθ=BINAsin90∘C\theta =BINAsin90^{\circ}
⇒C=BINAθ=0.01×10×10−3×100×2×10−40.05\Rightarrow C=\dfrac{BINA}{\theta }=\dfrac{0.01\times 10\times {10}^{-3}\times 100\times 2\times {10}^{-4}}{0.05}
=4×10−5N−m/rad.=4\times {10}^{-5}N-m/rad.
Therefore, x=4x=4.

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