NVAEasyJEE 2023Ampere's Law

JEE Physics 2023 Question with Solution

The magnetic intensity at the center of a long current carrying solenoid is found to be 1.6×103A/m1.6 \times 10^3\,A/m. If the number of turns is 88 per cm, then the current flowing through the solenoid is:

Answer

Correct answer:2

Step-by-step solution

Standard Method

Given: Magnetic intensity at the center of the solenoid is H=1.6×103A/mH = 1.6 \times 10^3 \, \text{A/m} and the number of turns is n=8turns/cm=800turns/mn = 8 \, \text{turns/cm} = 800 \, \text{turns/m}.

Find: The current II flowing through the solenoid.

For a long current carrying solenoid, magnetic intensity is given by

H=nIH = nI

So,

I=Hn=1.6×103800=2AI = \frac{H}{n} = \frac{1.6 \times 10^3}{800} = 2 \, \text{A}

Therefore, the current flowing through the solenoid is 2A2 \, \text{A}.

Common mistakes

  • Using n=8n = 8 directly without converting turns per centimetre to turns per metre is incorrect because HH is given in A/m\text{A/m}. Convert 8turns/cm8 \, \text{turns/cm} to 800turns/m800 \, \text{turns/m} before substitution.

  • Applying the formula for magnetic field BB instead of magnetic intensity HH leads to an unnecessary factor of μ0\mu_0. Here the given quantity is magnetic intensity, so use H=nIH = nI.

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