A moving coil galvanometer has 100 turns and each turn has an area of $$2.0 \text{ cm}^2$$. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is $$x \times 10^{-5}$$ N-m/rad. The value of $$x$$ is ______.

Find the torsional constant $$k = x \times 10^{-5}$$ N-m/rad of a galvanometer.

For a moving coil galvanometer, at equilibrium the magnetic torque equals the restoring torque:

$$ NBIA = k\theta $$

where $$N$$ = number of turns, $$B$$ = magnetic field, $$I$$ = current, $$A$$ = area of each turn, $$k$$ = torsional constant, $$\theta$$ = deflection angle.

$$N = 100$$, $$B = 0.01$$ T, $$I = 10$$ mA $$= 10 \times 10^{-3}$$ A, $$A = 2.0$$ cm$$^2$$ $$= 2.0 \times 10^{-4}$$ m$$^2$$, $$\theta = 0.05$$ rad.

$$ k = \frac{NBIA}{\theta} = \frac{100 \times 0.01 \times 10 \times 10^{-3} \times 2.0 \times 10^{-4}}{0.05} $$

Numerator: $$100 \times 0.01 = 1$$

$$1 \times 10 \times 10^{-3} = 10^{-2}$$

$$10^{-2} \times 2.0 \times 10^{-4} = 2.0 \times 10^{-6}$$

$$ k = \frac{2.0 \times 10^{-6}}{0.05} = \frac{2.0 \times 10^{-6}}{5 \times 10^{-2}} = 0.4 \times 10^{-4} = 4 \times 10^{-5} \text{ N-m/rad} $$

So $$x = 4$$.

The answer is 4.