Question

Question 4.10: Two moving coil meters, M1 and M2 have the following particulars: R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10 –3 m 2 , B1 = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10 –3 m 2 , B2 = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-169

Answer 1

current sensitivity of a moving coil galvanometer is defined as $C.S=\frac{\phi}{I}=\frac{BAN}{k}$
and also voltage sensitivity,$V.S=\frac{BAN}{kR}$
hence, it is clear that V.S = C.S/R

(i) ratio of current sensitivity ,
$\frac{C.S_2}{C.S_1}=\frac{N_2B_2A_2k}{N_1B_1A_1k}=\frac{N_2B_2A_2}{N_1B_1A_1}$
= (42 × 1.8 × 0.5 × 10^-3 )/(30 × 0.25 × 3.6 × 10^-3 )
= (21 × 1.8)/(15 × 1.8)
= 7/5

(ii) ratio of voltage sensitivity,
$\frac{V.S_2}{V.S_1}=\frac{C.S_2\times R_1}{C.S_1\times R_2}\\\\=\frac{7}{5}\times\frac{10}{14}=1$