A moving coil galvanometer has a coil with 175 turns and area 1 cm$$^2$$. It uses a torsion band of torsion constant $$10^{-6}$$ N m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in tesla) is approximately:

The coil of the moving-coil galvanometer reaches equilibrium when the magnetic torque trying to turn the coil is exactly balanced by the restoring (torsional) torque of the suspension.

First, we state the two torques.

Magnetic torque on a current-carrying coil is given by the formula $$\tau_{\text{mag}} = n\,I\,A\,B\,\sin\phi,$$ where $$n$$ is the number of turns, $$I$$ is the current, $$A$$ is the area of each turn, $$B$$ is the magnetic-field magnitude, and $$\phi$$ is the angle between the normal to the coil and the magnetic field.

The problem tells us that the magnetic field is parallel to the plane of the coil. Hence the normal to the coil is perpendicular to the field, so $$\phi = 90^\circ$$ and $$\sin\phi = \sin 90^\circ = 1.$$ Therefore, the magnetic torque reduces to $$\tau_{\text{mag}} = n\,I\,A\,B.$$

The restoring torque supplied by the torsion band obeys Hooke’s law for angular twist: $$\tau_{\text{rest}} = C\,\theta,$$ where $$C$$ is the torsion (restoring) constant and $$\theta$$ is the angular deflection in radians.

At equilibrium we have $$\tau_{\text{mag}} = \tau_{\text{rest}}.$$ So, $$n\,I\,A\,B = C\,\theta.$$

We need to isolate $$B$$:

$$B = \frac{C\,\theta}{n\,I\,A}.$$

Now we substitute every given quantity, converting each into SI units:

Number of turns: $$n = 175.$$

Current: $$I = 1\ \text{mA} = 1 \times 10^{-3}\ \text{A}.$$

Area of the coil: $$A = 1\ \text{cm}^2 = 1 \times 10^{-4}\ \text{m}^2.$$ (1 cm = 10⁻² m, so 1 cm² = 10⁻⁴ m².)

Torsion constant: $$C = 10^{-6}\ \text{N m rad}^{-1}.$$

Angular deflection: $$\theta = 1^\circ = \left(\frac{\pi}{180}\right)\ \text{rad} \approx 0.01745\ \text{rad}.$$

Substituting these numbers into the expression for $$B$$ gives

$$B = \frac{C\,\theta}{n\,I\,A} = \frac{(10^{-6})\,(0.01745)}{(175)\,(1 \times 10^{-3})\,(1 \times 10^{-4})}.$$

We now calculate the numerator:

$$C\,\theta = 10^{-6} \times 0.01745 = 1.745 \times 10^{-8}.$$

Next, the denominator:

$$n\,I\,A = 175 \times (1 \times 10^{-3}) \times (1 \times 10^{-4}) = 175 \times 10^{-7} = 1.75 \times 10^{-5}.$$

Dividing numerator by denominator:

$$B = \frac{1.745 \times 10^{-8}}{1.75 \times 10^{-5}}.$$

Because $$\frac{1.745}{1.75} \approx 0.9986,$$ we get

$$B \approx 0.9986 \times 10^{-3}\ \text{T} \approx 1 \times 10^{-3}\ \text{T}.$$

The value $$1 \times 10^{-3}\ \text{T}$$ corresponds to $$10^{-3}$$ tesla.

Hence, the correct answer is Option C.