In a galvanometer, the number of turns N is doubled (so resistance also doubles). The voltage sensitivity then:

1. Voltage sensitivity: $\dfrac{\phi}{V} = \dfrac{NAB}{k}\dfrac{1}{R}$. 2. Doubling N makes the numerator $N \to 2N$. 3. But the wire length, and hence resistance, also doubles: $R \to 2R$. 4. The factors cancel: $\dfrac{2N}{2R} = \dfrac{N}{R}$, so voltage sensitivity is unchanged. 5. (Current sensitivity, $\propto N$ alone, would double — a common trap.) _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.161_