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Magnetic Properties of Matter question

2023 · 8 Apr · Shift 1 · Q66
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  5. /2023 · 8 Apr · Shift 1 · Q66

Magnetic Properties of Matter question

2023 · 8 Apr · Shift 1 · Q66

JEE MainPhysicsMagnetic Properties of MatterNumerical+4 / −1
The magnetic intensity at the center of a long current carrying solenoid is found to be 1.6×103Am−11.6 \times 10^{3} \mathrm{Am}^{-1}1.6×103Am−1. If the number of turns is 8 per cm, then the current flowing through the solenoid is ‾\underline{\hspace{2cm}}​ A.
Numerical answer
View written solutionFree

Correct answer: 2

  1. Use the formula for magnetic intensity inside a long solenoid

For a long solenoid, H=nIH = nIH=nI where:

  • HHH = magnetic intensity
  • nnn = number of turns per unit length
  • III = current
  1. Given data

H=1.6×103 A m−1H = 1.6 \times 10^3\ \text{A m}^{-1}H=1.6×103 A m−1

Number of turns = 888 per cm.

Convert this into turns per metre: n=8×100=800 m−1n = 8 \times 100 = 800\ \text{m}^{-1}n=8×100=800 m−1

  1. Calculate the current

Using H=nIH = nIH=nI so, I=Hn=1.6×103800I = \frac{H}{n} = \frac{1.6 \times 10^3}{800}I=nH​=8001.6×103​

I=1600800=2 AI = \frac{1600}{800} = 2\ \text{A}I=8001600​=2 A

  1. Final answer

The current flowing through the solenoid is 2\boxed{2}2​

  1. Comparison with stored correct answer

Stored correct answer = 222

My derived answer also = 222, so they agree.

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