For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passed through it. If the torsional constant of suspension wire is $$4.0 \times 10^{-5}$$ N m rad$$^{-1}$$, the magnetic field is 0.01 T and the number of turns in the coil is 200, the area of each turn (in cm$$^2$$) is:

We need to find the area of each turn of a moving coil galvanometer.

Deflection $$\theta = 0.05$$ rad, current $$I = 10$$ mA = 0.01 A, torsional constant $$C = 4.0 \times 10^{-5}$$ N m/rad, magnetic field $$B = 0.01$$ T, number of turns $$N = 200$$.

We start by noting that

In a moving coil galvanometer, the magnetic torque on the coil ($$\tau_m = NBIA$$) is balanced by the restoring torque of the suspension wire ($$\tau_r = C\theta$$):

$$ NBIA = C\theta $$

where $$A$$ is the area of each turn.

Next,

$$ A = \frac{C\theta}{NBI} $$

$$ A = \frac{4.0 \times 10^{-5} \times 0.05}{200 \times 0.01 \times 0.01} $$

$$ A = \frac{2.0 \times 10^{-6}}{0.02} = 1.0 \times 10^{-4}\;\text{m}^2 $$

Converting to cm$$^2$$: $$1.0 \times 10^{-4}$$ m$$^2$$ = $$1.0$$ cm$$^2$$.

The area of each turn is 1.0 cm$$^2$$.

The correct answer is Option 2: 1.0.