Test Index

NCERT Class XII Chapter
Moving Charges and Magnetism
Questions With Solutions

© examsnet.com
Question : 10 of 28
Marks: +1, -0
Two moving coil meters, M1 and M2 have following particulars :
R1R_1 = 10 Ω, N1N_1 = 30, A1A_1 = 3.6 × 10−3 m210^{-3}\,\mathrm{m}^2, B1B_1 = 0.25 T
R2R_2 = 14 Ω, N2N_2 = 42, A2A_2 = 1.8 × 10−3 m210^{-3}\,\mathrm{m}^2, B2B_2 = 0.50 T.
(The spring constants are identical for two meters).
Determine ratio of (i) current sensitivity (ii) voltage sensitivity of M2M_2 and M1M_1.
Solution:  
Current sensitivity of a moving coil galvanometer is defined as
C.S. = ΦI\frac{\Phi}{I} = NABk\frac{NAB}{k}
and voltage sensitivity, V.S. =
NABkR\frac{NAB}{kR}
(i) Ratio of current sensitivity
C.S2C.S1\frac{C.S_2}{C.S_1} = N2B2A2kN1B1A1k\frac{N_2B_2A_2k}{N_1B_1A_1k} = 42×1.8×0.5×10−3×k30×0.25×03.6×10−3×k\frac{42\times1.8\times0.5\times10^{-3}\times k}{30\times0.25\times03.6\times10^{-3}\times k} = 1.4
(ii) Ratio of voltage sensitivity
V.S2V.S1\frac{V.S_2}{V.S_1} = C.S2×R1C.S1×R2\frac{C.S_2 \times R_1}{C.S_1 \times R_2} = 75×1014\frac{7}{5} \times \frac{10}{14} = 1
© examsnet.com
Go to Question:
Prev Question
Next Question