Moving Charges and Magnetism

Step: Find the magnetic force per unit length.
The force per unit length between two long parallel current-carrying wires is: 
\(\dfrac{F}{L}=\dfrac{\mu_0 i_1 i_2}{2\pi r}\)
\(\Rightarrow \dfrac{F}{L}\propto \dfrac{i_2 i_1}{r}\)
Hence, option (1) is the correct answer.
 

Step: Find the change in the voltage sensitivity.
The voltage sensitivity \((S_V)\) of a moving coil galvanometer is given by;
\(S_V=\frac{NAB}{KR}\)
where, \(N = \) Number of turns, \(B=\) Magnetic field, \(A=\) Area of the coil, \(K=\) Torsional constant, and \(R=\) Resistance of the coil.
The current sensitivity of the coil is given by;
\(S_I = \frac{\theta}{I}= \frac{NBA}{K}\)
The voltage sensitivity is related to current sensitivity by the expression as;
\(S_V = \frac{S_I}{R}\)
If the number of turns \(N\) increases by \(25\%\), it would increase the current sensitivity \((S_I)\) proportionally. However, the resistance \(R\) of the coil also increases proportionally to \(N\) (since more turns mean longer wire).
Since \(S_V = \frac{S_I}{R}\) and both \(S_I\)​ and \(R\) increase proportionally, the voltage sensitivity remains unchanged.
Therefore, the change in the voltage sensitivity is zero.
Hence, option (1) is the correct answer.