Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.10

NCERT Solutions Class 12 Physics Chapter 4 Question 10

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For moving coil meter M₁:

Resistance, R₁ = 10 QΩ

Number of turns, N₁ = 30

Area of cross-section, A₁ = 3.6 x 10⁻

^{3}m^{2}Magnetic field strength, B₁ = 0.25 T

Spring constant K₁ = K

For moving coil meter M₂:

Resistance, R₂= 14 Q

Number of turns, N₂= 42

Area of cross-section, A₂ = 1.8 x 10⁻

^{3}m^{2}Magnetic field strength, B₂ = 0.50 T

Spring constant, K₂ = K

Ans (a).Current sensitivity of M₁ is given by

Is₁ = N₁B₁A₁/K₁

and, current sensitivity of M₂ is given by

Is₂ = N₂B₂A₂/K₂

Therefore Ratio Is₂ /Is₁ = N₂B₂A₂ /N₁B₁A₁ (Since Spring constant, K₂ = K)

=> Is₂ /Is₁ = (42 x 0.5 x 1.8 x 10⁻

^{3}x K)/(K x 30 x 0.25 x 3.6 x 10⁻^{3}) = 1.4Hence ,the ratio of current sensitivity of M₂ to M₁ is 1.4

Ans (b).Voltage sensitivity for M₂ is given by

Vs₂ = N₂B₂A₂/ K₂ R₂

And Volatage sensitivity for M₁ is given by

Vs₁ = N₁B₁A₁/K₁R₁

Therefore Ratio Vs₂/ Vs₁ = N₂B₂A₂ K₁R₁/ K₂ R₂ N₁B₁A₁

Vs₂/ Vs₁ = (42 x 0.5 x 1.8 x 10⁻³ x 10 x K)/(K x 14 x 30 x 0.25 x 3.6 x 10⁻³ )

=> Vs₂/ Vs₁ = 1

Hence ,the ratio of voltage sensitivity of M₂ to M₁ is 1.