Question

The unit of reduction factor of tangent galvanometer:
(A)Ampere
(B)Gauss
(C)Radian
(D)No Units

Answer 1

** The concept involved here is that of a moving coil galvanometer, a sensitive instrument used to measure electric currents. A tangent galvanometer is a low current measuring device whose working principle is the tangent law of magnetism.
Formula Used:
$I=K\tan \theta$

Complete answer:
The current passing through the tangent galvanometer is given by the following equation:
$I=K\tan \theta$
Here,
K is called the reduction factor of the tangent galvanometer.
In the above equation, the unit of current (I) is Ampere. As we know tangent of an angle is dimensionless, so according to dimensional analysis, we conclude that K will have the same dimensions as that of current (I). Hence, the reduction factor (K) will have the unit of current.
The unit of reduction factor of a tangent galvanometer is:(A)Ampere.

Additional Information: The tangent law of magnetism states that the tangent of angle of the compass needle in a galvanometer is proportional to the ratio of the strengths of the two perpendicular magnetic fields.
The galvanometer is oriented in such a way that the plane of the coil of the galvanometer is aligned parallel to the horizontal component of Earth’s magnetic field. When an electric field is made to flow through the galvanometer, another magnetic field (B) is created.
$B=\dfrac{{{\mu }_{0}}nI}{r}$ $........(1)$
Here,
It is the current passed through the coil of the galvanometer.
n is the no of turns in the coil.
r is the radius of the coil.
The two magnetic fields add according to vector addition and the current in the coil causes the compass needle to rotate by an angle of theta.
From the tangent law, we have
$B={{B}_{H}}\tan \theta$
$\Rightarrow \theta ={{\tan }^{-1}}\dfrac{B}{{{B}_{H}}}$ $.......(2)$
Equating (1) and (2)
$\dfrac{{{\mu }_{0}}nI}{r}={{B}_{H}}\tan \theta$
$\begin{aligned} & \Rightarrow I=\dfrac{2r{{B}_{H}}}{{{\mu }_{0}}n}\tan \theta \\\ & \Rightarrow I=K\tan \theta \\\ & Here, \\\ \end{aligned}$

$K=\dfrac{2r{{B}_{H}}}{{{\mu }_{0}}n}$
Thus, we have derived the equation for reduction factor (K).

Note:
The reduction factor of a galvanometer is given by the amount of current required to pass through the galvanometer producing a specific deflection. It has the same units as that of electric current.