Step 1 : Understanding the Question:
The topic of this question is Magnetic Effects of Current, specifically the design and operation of a Moving Coil Galvanometer (MCG). A galvanometer is an instrument used to detect and measure small electric currents. One of its critical design features is the use of a "radial magnetic field" created by concave magnetic poles and a soft iron core. The question asks why this specific field shape is necessary for the instrument's functionality.
Step 2 : Key Formulas and approach:
The approach involves analyzing the torque acting on a current-carrying loop in a magnetic field:
1. Deflecting Torque: $\tau = NIAB \sin \theta$
2. Restoring Torque: $\tau = k \alpha$
3. Goal: We want a linear relationship between the angular deflection ($\alpha$) and the current ($I$), which means we want $\alpha \propto I$.
Where $N$ is turns, $I$ is current, $A$ is area, $B$ is magnetic field, $\theta$ is the angle between the area vector and $B$, and $k$ is the spring constant.
Step 3 : Detailed Explanation:
In a standard uniform magnetic field, the torque depends on the sine of the angle $\theta$. As the coil rotates, $\theta$ changes, which means the torque would change even if the current $I$ stayed the same. This would lead to a non-linear (logarithmic or sinusoidal) scale.
To make the galvanometer useful, we need a "linear scale" where a doubling of current results in exactly a doubling of the needle's deflection.
By making the magnetic field "radial," the magnetic field lines are always parallel to the plane of the coil (or perpendicular to the area vector) regardless of how much the coil has rotated.
This ensures that the angle $\theta$ between the magnetic field and the normal to the coil is always $90^\circ$. Since $\sin 90^\circ = 1$, the torque formula simplifies to $\tau = NIAB$.
At equilibrium, the deflecting torque equals the restoring torque of the spring: $NIAB = k \alpha$.
Rearranging for deflection, we get $\alpha = (\frac{NAB}{k}) I$. Since everything in the parentheses is constant, $\alpha \propto I$.
This allows the manufacturer to mark the scale with equal divisions for equal current increments.
Step 4 : Final Answer:
The radial magnetic field ensures a constant torque by keeping $\sin \theta = 1$, which makes the deflection directly proportional to the current. Thus, the correct option is (C).