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Electromagnetic Induction question

2008 · Shift 0 · Q63
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  5. /2008 · Shift 0 · Q63

Electromagnetic Induction question

2008 · Shift 0 · Q63

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A=10  cm2A=10\,\,c{m^2}A=10cm2 and length =20cm=20cm=20cm. If one of the solenoid has 300300300 turns and the other 400400400 turns, their mutual inductance is (μ0=4π×10−7 Tm A−1)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,Tm\,{A^{ - 1}}} \right)(μ0​=4π×10−7TmA−1)
  1. A
    2.4π×10−5H2.4\pi \times {10^{ - 5}}H2.4π×10−5H
  2. B
    4.8π×10−4H4.8\pi \times {10^{ - 4}}H4.8π×10−4H
  3. C
    4.8π×10−5H4.8\pi \times {10^{ - 5}}H4.8π×10−5H
  4. D
    2.4π×10−4H2.4\pi \times {10^{ - 4}}H2.4π×10−4H
View written solutionFree

Correct answer: D

  1. Formula for mutual inductance of two long coaxial solenoids

For two coaxial solenoids of the same length and common cross-sectional area,

M=μ0N1N2AlM = \mu_0 \frac{N_1 N_2 A}{l}M=μ0​lN1​N2​A​

where:

  • N1=300N_1 = 300N1​=300
  • N2=400N_2 = 400N2​=400
  • A=10 cm2=10×10−4 m2=10−3 m2A = 10\,\text{cm}^2 = 10 \times 10^{-4}\,\text{m}^2 = 10^{-3}\,\text{m}^2A=10cm2=10×10−4m2=10−3m2
  • l=20 cm=0.2 ml = 20\,\text{cm} = 0.2\,\text{m}l=20cm=0.2m
  • μ0=4π×10−7 H/m\mu_0 = 4\pi \times 10^{-7}\,\text{H/m}μ0​=4π×10−7H/m
  1. Substitute the values

M=4π×10−7×300×400×10−30.2M = 4\pi \times 10^{-7} \times \frac{300 \times 400 \times 10^{-3}}{0.2}M=4π×10−7×0.2300×400×10−3​

  1. Simplify step-by-step

First,

300×400=120000=1.2×105300 \times 400 = 120000 = 1.2 \times 10^5300×400=120000=1.2×105

Then,

10−30.2=10−32×10−1=5×10−3\frac{10^{-3}}{0.2} = \frac{10^{-3}}{2 \times 10^{-1}} = 5 \times 10^{-3}0.210−3​=2×10−110−3​=5×10−3

So,

300×400×10−30.2=120000×5×10−3=600\frac{300 \times 400 \times 10^{-3}}{0.2} = 120000 \times 5 \times 10^{-3} = 6000.2300×400×10−3​=120000×5×10−3=600

Hence,

M=4π×10−7×600M = 4\pi \times 10^{-7} \times 600M=4π×10−7×600

M=2400π×10−7M = 2400\pi \times 10^{-7}M=2400π×10−7

M=2.4π×10−4 HM = 2.4\pi \times 10^{-4}\,\text{H}M=2.4π×10−4H

  1. Match with the options

This corresponds to:

Option D: 2.4π×10−4 H2.4\pi \times 10^{-4}\,\text{H}2.4π×10−4H

  1. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

They match.

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