Question

A tangent galvanometer is connected directly to an ideal battery. If the number of turns in the coil is doubled, the deflection will

• A. increase
• B. decrease
• C. remain unchanged
• D. any of these

Answer 1

The correct option is C remain unchanged

Current passing through the coil of a tangent galvanometer is given by,
$$i=\left(\frac{2a{B}_H}{{\mu }_{0}N}\right)\tan \theta$$
Here,
$a\to$ Radius of the coil
$B_H\to$ Horizontal component of the magnetic field
$N\to$ Number of turns of the coil
$\theta \to$ Angle of deflection

If the potential difference across the coil is $V$ and the coil's resistance is $R$, then

$i=\frac{V}{R}=\left(\frac{2a{B}_H}{{\mu }_{0}N}\right)\tan \theta$

$\Rightarrow \tan \theta =\frac{{\mu }_{0}NV}{2a{B}_{H}R}............\left(1\right)$

When the number of turns are doubled, the resistance of the coil is also doubled, so the new deflection

$\Rightarrow \tan {\theta }^{\prime }=\frac{{\mu }_0\left(2N\right)V}{2a{B}_H\left(2R\right)}=\frac{{\mu }_{0}NV}{2a{B}_{H}R}.........\left(2\right)$

From eqs. $\left(1\right)$ and $\left(2\right)$,

${\theta }^{\prime }=\theta$

Hence, option $\left(C\right)$ is the correct answer.