Question

Reduction factor of a tangent galvanometer is related to number of turns as :

• A. $n$
• B. $n^2$
• C. $n^{-1/2.}$
• D. $n^{-1}$

Answer 1

Answer: (D)

$n^{-1}$

From tangent law we have $B=B_H\tan \theta$,

$B$ is the second magnetic field created by the flow of current($I$), which is given by $B=\frac{{\mu }_{0}nI}{2r}$.

The reduction factor, $K=\frac{I}{\tan \theta }$, therefore $K=p\frac{1}{n}$ where $p$ is a constant.