JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(10^{-3}\)
- B \(10^{-1}\)
- C \(10^{-4}\)
- D \(10^{-2}\)
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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