In a moving coil galvanometer, two moving coils M₁ and M₂ have the following particulars :
R₁=5 Ω, N₁=15, A₁=3.6 × 10⁻³ m², B₁=0.25 T ; R₂=7 Ω, N₂=21, A₂=1.8 × 10⁻³ m², B₂=0.50 T
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M₁ and M₂ ?

Solution

Physics sir · 3 steps

Voltage sensitivity is NAB divided by kR, so work out NAB/R for each coil and compare.

1. Recall the formula. Current sensitivity is deflection per unit current, θ/I = NAB/k. Since I = V/R, deflection per unit voltage picks up an extra 1/R.

   θ/V = NAB/kR

2. Cancel k. The torsional constant k is the same for both springs, so the ratio depends only on NAB/R for each coil.

   Coil	NAB (in 10⁻³ units)	R (Ω)	NAB/R (in 10⁻³ units)
   M₁	15 × 3.6 × 0.25 = 13.5	5	13.5/5 = 2.7
   M₂	21 × 1.8 × 0.50 = 18.9	7	18.9/7 = 2.7

3. So. Both coils give NAB/R = 2.7 × 10⁻³ in SI units, so the voltage sensitivities are in the ratio 1 : 1, option (a).

The idea to keep. Current sensitivity is NAB/k. Voltage sensitivity is NAB/(kR). Always check whether the coil resistance belongs in the comparison before you compare anything.

Answer from the NTA final answer key. Solution by PrepWiser.