Alternate Answer
Voltage sensitivity is the deflection per unit voltage.Alternate Answer
$\frac{\phi}{\text{V}} = \bigg(\frac{\text{NBA}}{\text{k}}\bigg) \frac{\text{I}}{\text{V}} = \bigg(\frac{\text{NAB}}{\text{k}}\bigg)\frac{1}{\text{R}}$ Justification: Increasing the current sensitivity may not necessarily increase the voltage sensitivity. If N→2N, i.e., we double the number of turns, then $\frac{\phi}{\text{I}}\rightarrow2\frac{\phi}{\text{I}}$ Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. N→2N, and R→2R, thus the voltage sensitivity, remains unchanged. $\frac{\phi}{\text{V}}\rightarrow\frac{\phi}{\text{V}}.$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{F = qE, F = qvB,}$ and $\text{B}=\frac{\mu_0\text{I}}{2\pi\alpha};$
where F is force, q is charge, v is speed, I is current, and a is distance.