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
Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10⁻³m², B₁ = 0.25 T R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 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.
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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 m2
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 m2
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 ) = 1.4
Hence ,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.