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In a moving coil galvanometer the magnetic field is made radial chiefly so that:

AThe deflection stays proportional to current
BThe coil resistance is greatly reduced
CThe spring constant becomes much larger
DThe current required is greatly increased
Answer & Solution
Correct answer: A. The deflection stays proportional to current
1. The torque on the coil is $\tau = NIAB\sin\theta$ in general. 2. A radial field keeps the plane of the coil always along B, so $\theta = 90^\circ$ and $\sin\theta = 1$. 3. The torque then reduces to $NIAB$, independent of deflection angle. 4. Balancing with $k\phi$ gives $\phi \propto I$, a linear, easily-read scale. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.159_
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