Torque on a moving coil galvanometer

At equilibrium, $\text{magnetic torque} = \text{restoring torque}$

Magnetic torque on the coil: $\tau_m = N B I A$

Restoring torque due to suspension wire: $\tau_r = C\,\theta$

where $C$ is the torsional constant.

So, $N B I A = C\theta$

Hence, $C = \frac{N B I A}{\theta}$

Substitute the given values

• Number of turns: $N = 100$
• Area of each turn: $A = 2.0\,\text{cm}^2 = 2.0 \times 10^{-4}\,\text{m}^2$
• Magnetic field: $B = 0.01\,\text{T}$
• Current: $I = 10\,\text{mA} = 10 \times 10^{-3} = 0.01\,\text{A}$
• Deflection: $\theta = 0.05\,\text{rad}$

Therefore, $C = \frac{100 \times 0.01 \times 0.01 \times 2.0\times 10^{-4}}{0.05}$

Calculate step-by-step

First, $100 \times 0.01 = 1$

Then, $1 \times 0.01 = 0.01$

Next, $0.01 \times 2.0\times 10^{-4} = 2.0\times 10^{-6}$

Now divide by $0.05$: $C = \frac{2.0\times 10^{-6}}{0.05} = 4.0\times 10^{-5}\,\text{N m/rad}$

Match with the required form

Given, $C = x \times 10^{-5}\,\text{N m/rad}$

Comparing, $x = 4$

Final answer

$\boxed{4}$