The error in measuring the current with a tangent galvanometer is minimum when the deflection is about:
A. $0^\circ $
B. $30^\circ $
C. $45^\circ $
D. $60^\circ $

Hint: A galvanometer is a device that detects and measures the magnitude of small electric currents. A key component of a galvanometer, the movement of a coil or magnetic needle in a magnetic field normally provides information about the current and its strength.

Formula used:
The expression of current in case of tangent galvanometer is given as,
$i = k\tan \phi $
Here, $k$ is the constant and $\phi$ is the deflecting angle.

Complete step by step solution:
As we know that the tangent galvanometers were the first instruments used to measure small electric currents It is composed of an insulated copper wire coil wound around a nonmagnetic circular frame and its operation is based on magnetism's tangent law.

When a current is passed through the centre of the circular coil in a direction perpendicular to the plane of the coil, a magnetic field is produced. And the horizontal component of the earth's magnetic field is pointed in the plane of the coil by the way, the TG is set up. So, in the case of tangent galvanometer as:
$i = k\tan \phi \,\,...(1)$

Differentiating the above equation with respect to $\phi $, then we have:
$\dfrac{{di}}{{d\phi }} = \dfrac{d}{{d\phi }}(k\tan \phi ) \\$
$\Rightarrow \dfrac{{di}}{{d\phi }} = k{\sec ^2}\phi \\$
$\Rightarrow di = k{\sec ^2}\phi d\phi \,\,....(2) \\$
Now, divide the equation $(1)$ and $(2)$, then we obtain:
$\dfrac{{di}}{i} = \dfrac{{k{{\sec }^2}\phi d\phi }}{{k\tan \phi }} \\$
$\Rightarrow \dfrac{{di}}{i} = \dfrac{{d\phi }}{{\sin \phi \cos \phi }} \\$
$\Rightarrow \dfrac{{di}}{i} = \dfrac{{2d\phi }}{{\sin 2\phi }} \\$
Which implies, $\sin 2\phi = 1$
$2\phi = 90^\circ \\$
$\therefore \phi = 45^\circ $
Therefore, the error in the measurement will be least when $\phi = 45^\circ $.

Thus, the correct option is C.

Note: It should be noted that one field is caused by the horizontal component of the earth's magnetic field, while the other is produced by passing current through the tangent galvanometer's coil. The needle comes to rest making an angle with ${B_H}$ due to the action of two magnetic fields, such that $B = {B_H}$.