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Class 12
Physics :-NCERT Solutions - Moving Charges and Magnetism.

Question 10
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A= 3.6 x 10-3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 x 10-3 m2, 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.


Answer:

For moving coil meter M1:

Resistance, R1 = 10 Ω

Number of turns, N1 = 30

Area of cross-section, A= 3.6 × 10-3 m2

Magnetic field strength, B1 = 0.25 T

Spring constant K1 = K

For moving coil meter M2:

Resistance, R2 = 14 Ω

Number of turns, N2 = 42

Area of cross-section, A2 = 1.8 × 10-3 m2

Magnetic field strength, B2 = 0.50 T

Spring constant, K2 = K

(a) Current sensitivity of M1 is given as:

And, current sensitivity of M2 is given as:

 

so,  Ratio 

Hence, the ratio of current sensitivity of M2 to M1 is 1.4.

(b) Voltage sensitivity for M2 is given as:

And, voltage sensitivity for M1 is given as: