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Magnetics question

2025 · 2 Apr · Shift 2 · Q52
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  5. /2025 · 2 Apr · Shift 2 · Q52

Magnetics question

2025 · 2 Apr · Shift 2 · Q52

JEE MainPhysicsMagneticsMCQ+4 / −1
In a moving coil galvanometer, two moving coils M1\mathrm{M}_1M1​ and M2\mathrm{M}_2M2​ have the following particulars : R1=5Ω, N1=15, A1=3.6×10−3 m2, B1=0.25 TR2=7Ω, N2=21, A2=1.8×10−3 m2, B2=0.50 T\begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \\ & \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T} \end{aligned}​R1​=5Ω, N1​=15, A1​=3.6×10−3 m2, B1​=0.25 TR2​=7Ω, N2​=21, A2​=1.8×10−3 m2, B2​=0.50 T​ Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M1M_1M1​ and M2M_2M2​ ?
  1. A
    1:11: 11:1
  2. B
    1:31: 31:3
  3. C
    1:21: 21:2
  4. D
    1:41: 41:4
View written solutionFree

Correct answer: A

  1. Voltage sensitivity of a moving coil galvanometer

For a moving coil galvanometer,

τ=NBIA\tau = N B I Aτ=NBIA

and at equilibrium,

Cθ=NBIAC\theta = N B I ACθ=NBIA

where CCC is the torsional constant.

So,

θI=NBAC\frac{\theta}{I} = \frac{N B A}{C}Iθ​=CNBA​

This is the current sensitivity.

Now, voltage sensitivity is

θV=θI⋅IV=NBAC⋅1R\frac{\theta}{V} = \frac{\theta}{I} \cdot \frac{I}{V} = \frac{N B A}{C} \cdot \frac{1}{R}Vθ​=Iθ​⋅VI​=CNBA​⋅R1​

Thus,

Voltage sensitivity∝NBAR\text{Voltage sensitivity} \propto \frac{N B A}{R}Voltage sensitivity∝RNBA​

(since torsional constant CCC is same for both coils).

  1. For coil M1M_1M1​

Given:

R1=5 Ω,N1=15,A1=3.6×10−3 m2,B1=0.25 TR_1=5\,\Omega,\quad N_1=15,\quad A_1=3.6\times 10^{-3}\,\text{m}^2,\quad B_1=0.25\,\text{T}R1​=5Ω,N1​=15,A1​=3.6×10−3m2,B1​=0.25T

So,

N1B1A1R1=15×0.25×3.6×10−35\frac{N_1 B_1 A_1}{R_1} = \frac{15 \times 0.25 \times 3.6\times 10^{-3}}{5}R1​N1​B1​A1​​=515×0.25×3.6×10−3​

First,

15×0.25=3.7515\times 0.25 = 3.7515×0.25=3.75 3.75×3.6×10−3=13.5×10−33.75 \times 3.6\times 10^{-3} = 13.5\times 10^{-3}3.75×3.6×10−3=13.5×10−3

Now divide by 5:

13.5×10−35=2.7×10−3\frac{13.5\times 10^{-3}}{5} = 2.7\times 10^{-3}513.5×10−3​=2.7×10−3
  1. For coil M2M_2M2​

Given:

R2=7 Ω,N2=21,A2=1.8×10−3 m2,B2=0.50 TR_2=7\,\Omega,\quad N_2=21,\quad A_2=1.8\times 10^{-3}\,\text{m}^2,\quad B_2=0.50\,\text{T}R2​=7Ω,N2​=21,A2​=1.8×10−3m2,B2​=0.50T

So,

N2B2A2R2=21×0.50×1.8×10−37\frac{N_2 B_2 A_2}{R_2} = \frac{21 \times 0.50 \times 1.8\times 10^{-3}}{7}R2​N2​B2​A2​​=721×0.50×1.8×10−3​

First,

21×0.50=10.521\times 0.50 = 10.521×0.50=10.5 10.5×1.8×10−3=18.9×10−310.5 \times 1.8\times 10^{-3} = 18.9\times 10^{-3}10.5×1.8×10−3=18.9×10−3

Now divide by 7:

18.9×10−37=2.7×10−3\frac{18.9\times 10^{-3}}{7} = 2.7\times 10^{-3}718.9×10−3​=2.7×10−3
  1. Ratio of voltage sensitivities
SV1:SV2=2.7×10−3:2.7×10−3=1:1S_{V1} : S_{V2} = 2.7\times 10^{-3} : 2.7\times 10^{-3} = 1:1SV1​:SV2​=2.7×10−3:2.7×10−3=1:1
  1. Option check

The correct option is:

A: 1:1\boxed{\text{A: } 1:1}A: 1:1​
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