JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A galvanometer coil has \(500\) turns and each turn has an average area of \(3 \times 10^{-4}\, m ^{2}\). If a torque of \(1.5\,Nm\) is required to keep this coil parallel to magnetic field when a current of \(0.5\, A\) is flowing through it, the strength of the field (in \(T )\) is
- A \(15\)
- B \(20\)
- C \(35\)
- D \(30\)
Answer & Solution
Correct Answer
(B) \(20\)
Step-by-step Solution
Detailed explanation
\(\vec{\tau}=\vec{m} \times \vec{B}\) \(\tau=N I \times A \times B\) \(105=500 \times 3 \times 10^{-4} \times \frac{1}{2} \times B\) \(B=20\)
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