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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2019
  • Medium
  • MCQ

A moving coil galvanometer has a coil with \(175\, turns\) and area \(1\, cm^2\) It uses a torsion band of torsion constant \(10^{-6}\, N\, -m/rad\). The coil is placed in a magnetic field \(B\) parallel to its plane. The coil deflects by \(10\) for a current of \(1\, mA\). The value of \(B\) (in Tesla) is approximately

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Answer & Solution

Correct Answer

(A) \(10^{-3}\)

Step-by-step Solution

Detailed explanation

\(\tau=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}\)

\(\mathrm{C} \theta=\mathrm{i} \mathrm{N} \mathrm{A} \mathrm{B}\)

\(10^{-6} \times \frac{\pi}{180}=10^{-3} \times 10^{-4} \times 175 \times B\)

\(B=10^{-3}\) \(Tesla\)

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