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Magnetics question

2025 · 24 Jan · Shift 2 · Q75
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Magnetics question

2025 · 24 Jan · Shift 2 · Q75

JEE MainPhysicsMagneticsNumerical+4 / −1
A tightly wound long solenoid carries a current of 1.5 A . An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns . The number of turns per metre in the solenoid is ‾\underline{\hspace{2cm}}​. [Take mass of electron me=9×10−31 kg\mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg}me​=9×10−31 kg, charge of electron ∣qe∣=1.6×10−19C\left|\mathrm{q}_{\mathrm{e}}\right|=1.6 \times 10^{-19} \mathrm{C}∣qe​∣=1.6×10−19C, μ0=4π×10−7 N A2,1 ns=10−9 s]\left.\mu_0=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^2}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right]μ0​=4π×10−7 A2 N​,1 ns=10−9 s]
Numerical answer
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Correct answer: 250

  1. Magnetic field inside a long solenoid

For a tightly wound long solenoid, B=μ0nIB = \mu_0 n IB=μ0​nI where:

  • BBB = magnetic field inside solenoid
  • nnn = number of turns per metre
  • I=1.5 AI = 1.5\,\text{A}I=1.5A
  1. Time period of electron in magnetic field

An electron moving perpendicular to a uniform magnetic field executes circular motion with cyclotron time period: T=2πmqBT = \frac{2\pi m}{qB}T=qB2πm​ Using magnitude of charge, B=2πmqTB = \frac{2\pi m}{qT}B=qT2πm​

Given:

  • m=9×10−31 kgm = 9 \times 10^{-31}\,\text{kg}m=9×10−31kg
  • q=1.6×10−19 Cq = 1.6 \times 10^{-19}\,\text{C}q=1.6×10−19C
  • T=75 ns=75×10−9 sT = 75\,\text{ns} = 75 \times 10^{-9}\,\text{s}T=75ns=75×10−9s

So, B=2π(9×10−31)(1.6×10−19)(75×10−9)B = \frac{2\pi (9 \times 10^{-31})}{(1.6 \times 10^{-19})(75 \times 10^{-9})}B=(1.6×10−19)(75×10−9)2π(9×10−31)​

Now simplify the denominator: 1.6×75=1201.6 \times 75 = 1201.6×75=120 (1.6×10−19)(75×10−9)=120×10−28=1.2×10−26 (1.6 \times 10^{-19})(75 \times 10^{-9}) = 120 \times 10^{-28} = 1.2 \times 10^{-26}(1.6×10−19)(75×10−9)=120×10−28=1.2×10−26

Thus, B=18π×10−311.2×10−26B = \frac{18\pi \times 10^{-31}}{1.2 \times 10^{-26}}B=1.2×10−2618π×10−31​ B=15π×10−5 TB = 15\pi \times 10^{-5}\,\text{T}B=15π×10−5T

Using π≈3.14\pi \approx 3.14π≈3.14, B≈47.1×10−5=4.71×10−4 TB \approx 47.1 \times 10^{-5} = 4.71 \times 10^{-4}\,\text{T}B≈47.1×10−5=4.71×10−4T

  1. Equate with solenoid field

μ0nI=B\mu_0 n I = Bμ0​nI=B n=Bμ0In = \frac{B}{\mu_0 I}n=μ0​IB​

Substitute values: n=15π×10−5(4π×10−7)(1.5)n = \frac{15\pi \times 10^{-5}}{(4\pi \times 10^{-7})(1.5)}n=(4π×10−7)(1.5)15π×10−5​

Cancel π\piπ: n=15×10−56×10−7n = \frac{15 \times 10^{-5}}{6 \times 10^{-7}}n=6×10−715×10−5​ n=156×102n = \frac{15}{6} \times 10^2n=615​×102 n=2.5×102=250n = 2.5 \times 10^2 = 250n=2.5×102=250

  1. Final answer

The number of turns per metre is 250\boxed{250}250​

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