Question

Two tangent galvanometers $\text{A}$ and $\text{B}$ have coils of radii $8\text{cm}$ and $16\text{cm}$ respectively and resistances $8\Omega$ each. They are connected in parallel with a cell of $\text{emf}4\text{V}$ and negligible internal resistance. The deflections produced in the tangent galvanometer $\text{A}$ and $\text{B}$ are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. If $\text{A}$ has two turns then $\text{B}$ must have .

Answer 1

As we know, that current in the tangent galvanometer is

$i=K\tan \theta$

Where,
Reduction factor, $K=\frac{2B_{H}a}{{\mu }_{0}n}$ and $a$ is the radius of the coil.

So, for constant $B_H\&{\mu }_0$,

$i\propto \frac{a\tan \theta }{n}$

$\therefore \frac{i_A}{i_B}=\frac{a_A\tan {\theta }_A}{a_B\tan {\theta }_B}\times \frac{n_B}{n_A}$

Given:
$a_A=8\text{cm};a_B=16\text{cm}$
${\theta }_A={30}^{\circ };{\theta }_B={60}^{\circ };n_A=2$

$\frac{i_A}{i_B}=\frac{8\tan {30}^{\circ }}{16\tan {60}^{\circ }}\times \frac{n_B}{2}=\frac{n_B}{12}$

Since, the resistance of the coils are same, and they are connected across the same cell, hence the current will be same in both the coils.

$\therefore i_A=i_B$

$\Rightarrow 1=\frac{n_B}{12}$

$\therefore n_B=12$

Accepted answers: $12\text{or}12.0\text{or}12.00$