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 :
- A \(2\)
- B \(1\)
- C \(1.5\)
- D \(0.5\)
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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