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Magnetics question

2025 · 29 Jan · Shift 2 · Q75
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Magnetics question

2025 · 29 Jan · Shift 2 · Q75

JEE MainPhysicsMagneticsNumerical+4 / −1
The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9×10−4 Tesla2.9 \times 10^{-4} ~\mathrm{Tesla}2.9×10−4 Tesla. If the solenoid carries a current of 0.29 A , then the length of the solenoid is ‾πcm\underline{\hspace{2cm}}\pi \mathrm{cm}​πcm.
Numerical answer
View written solutionFree

Correct answer: 8

  1. Magnetic field inside a solenoid

For a long solenoid,

B=μ0nI=μ0NLIB = \mu_0 n I = \mu_0 \frac{N}{L} IB=μ0​nI=μ0​LN​I

where:

  • B=2.9×10−4 TB = 2.9 \times 10^{-4}\,\text{T}B=2.9×10−4T
  • N=200N = 200N=200
  • I=0.29 AI = 0.29\,\text{A}I=0.29A
  • L=L =L= length of solenoid
  • μ0=4π×10−7 T m A−1\mu_0 = 4\pi \times 10^{-7}\,\text{T m A}^{-1}μ0​=4π×10−7T m A−1
  1. Solve for LLL

Rearranging,

L=μ0NIBL = \frac{\mu_0 N I}{B}L=Bμ0​NI​

Substitute the values:

L=(4π×10−7)(200)(0.29)2.9×10−4L = \frac{(4\pi \times 10^{-7})(200)(0.29)}{2.9 \times 10^{-4}}L=2.9×10−4(4π×10−7)(200)(0.29)​
  1. Simplify

First,

200×0.29=58200 \times 0.29 = 58200×0.29=58

So,

L=4π×10−7×582.9×10−4L = \frac{4\pi \times 10^{-7} \times 58}{2.9 \times 10^{-4}}L=2.9×10−44π×10−7×58​

Now,

582.9=20\frac{58}{2.9} = 202.958​=20

Hence,

L=4π×10−7×20×104L = 4\pi \times 10^{-7} \times 20 \times 10^4L=4π×10−7×20×104

Because

110−4=104\frac{1}{10^{-4}} = 10^410−41​=104

thus,

L=80π×10−3 mL = 80\pi \times 10^{-3}\,\text{m}L=80π×10−3m L=0.08π mL = 0.08\pi\,\text{m}L=0.08πm
  1. Convert to cm

Since 1 m=100 cm1\,\text{m} = 100\,\text{cm}1m=100cm,

L=0.08π×100=8π cmL = 0.08\pi \times 100 = 8\pi\,\text{cm}L=0.08π×100=8πcm

So the blank is:

8\boxed{8}8​
  1. Comparison with stored answer

Stored correct answer = 888

Derived answer = 888

They match.

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