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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

For a moving coil galvanometer, the deflection in the coil is \(0.05\,rad\) when a current of \(10\,mA\) is passed through it. If the torsional constant of suspension wire is \(4.0 \times 10^{-5}\,Nm\,rad ^{-1}\), the magnetic field is \(0.01\,T\) and the number of turns in the coil is \(200\),the area of each turn (in \(cm ^2\) ) is :

  1. A \(2\)
  2. B \(1\)
  3. C \(1.5\)
  4. D \(0.5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1\)

Step-by-step Solution

Detailed explanation

\(\tau= K \theta\) \(NiAB = K \theta\) \(A =\frac{ K \theta}{ NiB }=\frac{4 \times 10^{-5} \times 0.05}{200 \times 10 \times 10^{-3} \times 0.01}\) On solving \(A =10^{-4}\,m ^2=1\,cm ^2\)
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