Magnetic Effects & Magnetism

JEE Main level, single correct, medium

203 of 276

Question 203 of 276

For a moving coil galvanometer, the deflection in the coil is 0.05rad0.05rad when a current of 10mA10mA is passed through it. If the torsional constant of suspension wire is 4.0×10–5Nmrad−14.0\times {10}^{–5}Nm{rad}^{-1}, the magnetic field is 0.01T0.01T and the number of turns in the coil is 200200, the area of each turn (in cm2{cm}^{2} ) is :

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Answer: (B) 1.01.0

Given: θ=0.05rad,I=10mA,K=4×10−5Nmrad−1,B=0.01T,N=200\theta =0.05rad,I=10mA,K=4\times {10}^{-5}Nm{rad}^{-1},B=0.01T,N=200
The torsional constant of a wire used in a moving coil galvanometer is given by
K=NIABθK=\dfrac{NIAB}{\theta }, where, AA is area of each turn
So, A=KθNIB=4×10−5×0.05200×10×10−3×0.01A=\dfrac{K\theta }{NIB}=\dfrac{4\times {10}^{-5}\times 0.05}{200\times 10\times {10}^{-3}\times 0.01}
Thus, A=10−4m2=1cm2A={10}^{-4}{m}^{2}=1{cm}^{2}

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