This page focuses on the detailed Moving Charges and Magnetism question answers for Class 12 Physics Moving Charges and Magnetism, addressing the question: 'Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 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.'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.

Question 10

Two moving coil meters, M_{1} and M_{2} have the following particulars:

R_{1} = 10 Ω, N_{1} = 30,

A_{1 }= 3.6 x 10^{-3} m2, B1 = 0.25 T

R_{2} = 14 Ω, N_{2} = 42,

A_{2} = 1.8 x 10-3 m2, B_{2} = 0.50 T

(The spring constants are identical for the two meters).

Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M_{2} and M_{1}.

Answer

For moving coil meter M_{1}:

Resistance, R_{1} = 10 Ω

Number of turns, N_{1} = 30

Area of cross-section, A_{1 }= 3.6 × 10^{-3} m^{2}

Magnetic field strength, B_{1} = 0.25 T

Spring constant K_{1} = K

For moving coil meter M_{2}:

Resistance, R_{2} = 14 Ω

Number of turns, N_{2} = 42

Area of cross-section, A_{2} = 1.8 × 10_{-3} m^{2}

Magnetic field strength, B_{2} = 0.50 T

Spring constant, K_{2} = K

(a) Current sensitivity of M_{1} is given as:

And, current sensitivity of M_{2} is given as:

so, Ratio

Hence, the ratio of current sensitivity of M_{2} to M_{1} is 1.4.

(b) Voltage sensitivity for M_{2} is given as:

And, voltage sensitivity for M_{1} is given as:

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Nupur
2019-06-05 09:47:20

Not clear can't understand

Aditya Bhushan
2018-11-08 11:26:26

Not clear

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