Question:
A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant \(10^{-6}\) N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by \(1^\circ\) for a current of 1 mA. The value of B (in Tesla) is approximately :-
A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant \(10^{-6}\) N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by \(1^\circ\) for a current of 1 mA. The value of B (in Tesla) is approximately :-
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In equilibrium the magnetic torque NIAB (field in the plane of the coil) equals the restoring torque of the torsion band, C\(\theta\). Convert 1 degree to radians and the area to m2.
Updated On: Nov 5, 2026
- \(10^{-4}\)
- \(10^{-2}\)
- \(10^{-1}\)
- \(10^{-3}\)
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The Correct Option is D
Solution and Explanation
Magnetic torque (B in the plane of the coil) = NIAB; restoring torque = C\(\theta\).
NIAB = C\(\theta\) \(\Rightarrow\) \(B=\dfrac{C\theta}{NIA}\).
N = 175, I = 1 mA = \(10^{-3}\) A, A = 1 cm2 = \(10^{-4}\) m2, C = \(10^{-6}\) N-m/rad, \(\theta = 1^\circ = \dfrac{\pi}{180}=1.745\times10^{-2}\) rad.
\(NIA = 175\times10^{-3}\times10^{-4}=1.75\times10^{-5}\) A m2.
\(C\theta = 10^{-6}\times1.745\times10^{-2}=1.745\times10^{-8}\) N m.
\(B=\dfrac{1.745\times10^{-8}}{1.75\times10^{-5}}\approx 1.0\times10^{-3}\) T.
Answer: (4) \(10^{-3}\)
NIAB = C\(\theta\) \(\Rightarrow\) \(B=\dfrac{C\theta}{NIA}\).
N = 175, I = 1 mA = \(10^{-3}\) A, A = 1 cm2 = \(10^{-4}\) m2, C = \(10^{-6}\) N-m/rad, \(\theta = 1^\circ = \dfrac{\pi}{180}=1.745\times10^{-2}\) rad.
\(NIA = 175\times10^{-3}\times10^{-4}=1.75\times10^{-5}\) A m2.
\(C\theta = 10^{-6}\times1.745\times10^{-2}=1.745\times10^{-8}\) N m.
\(B=\dfrac{1.745\times10^{-8}}{1.75\times10^{-5}}\approx 1.0\times10^{-3}\) T.
Answer: (4) \(10^{-3}\)
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