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Magnetics question

2023 · 25 Jan · Shift 2 · Q63
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  5. /2023 · 25 Jan · Shift 2 · Q63

Magnetics question

2023 · 25 Jan · Shift 2 · Q63

JEE MainPhysicsMagneticsMCQ+4 / −1
For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is 4.0×10−5N m rad−14.0\times10^{-5}\mathrm{N~m~rad^{-1}}4.0×10−5N m rad−1, the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm 2^22) is :
  1. A
    1.5
  2. B
    2.0
  3. C
    0.5
  4. D
    1.0
View written solutionFree

Correct answer: D

  1. Use the torque balance for a moving coil galvanometer

At equilibrium, magnetic torque=restoring torque\text{magnetic torque} = \text{restoring torque}magnetic torque=restoring torque

So, NBIA=CθN B I A = C\thetaNBIA=Cθ where:

  • N=200N = 200N=200
  • B=0.01 TB = 0.01\,\text{T}B=0.01T
  • I=10 mA=0.01 AI = 10\,\text{mA} = 0.01\,\text{A}I=10mA=0.01A
  • A=A =A= area of each turn
  • C=4.0×10−5 N m rad−1C = 4.0\times 10^{-5}\,\text{N m rad}^{-1}C=4.0×10−5N m rad−1
  • θ=0.05 rad\theta = 0.05\,\text{rad}θ=0.05rad
  1. Substitute the values

200×0.01×0.01×A=4.0×10−5×0.05200 \times 0.01 \times 0.01 \times A = 4.0\times 10^{-5} \times 0.05200×0.01×0.01×A=4.0×10−5×0.05

Left side: 200×0.01×0.01=0.02200 \times 0.01 \times 0.01 = 0.02200×0.01×0.01=0.02

Right side: 4.0×10−5×0.05=2.0×10−64.0\times 10^{-5} \times 0.05 = 2.0\times 10^{-6}4.0×10−5×0.05=2.0×10−6

Thus, 0.02A=2.0×10−60.02A = 2.0\times 10^{-6}0.02A=2.0×10−6

  1. Solve for AAA

A=2.0×10−60.02=1.0×10−4 m2A = \frac{2.0\times 10^{-6}}{0.02} = 1.0\times 10^{-4}\,\text{m}^2A=0.022.0×10−6​=1.0×10−4m2

  1. Convert to cm2^22

Since, 1 m2=104 cm21\,\text{m}^2 = 10^4\,\text{cm}^21m2=104cm2

Therefore, A=1.0×10−4×104=1.0 cm2A = 1.0\times 10^{-4} \times 10^4 = 1.0\,\text{cm}^2A=1.0×10−4×104=1.0cm2

  1. Match with options

1.0 cm2\boxed{1.0\,\text{cm}^2}1.0cm2​

So the correct option is D.

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