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JEE Physics 2023 Question with Solution

A solenoid of 12001200 turns is wound uniformly in a single layer on a glass tube 2 m2 \, \text{m} long and 0.2 m0.2 \, \text{m} in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A2 \, \text{A} flows through it is:

  • A

    2.4×103 A m−12.4 \times 10^3 \, \text{A m}^{-1}

  • B

    1.2×103 A m−11.2 \times 10^3 \, \text{A m}^{-1}

  • C

    1 A m−11 \, \text{A m}^{-1}

  • D

    4.2×103 A m−14.2 \times 10^3 \, \text{A m}^{-1}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Total number of turns =1200= 1200, length of solenoid =2 m= 2 \, \text{m}, current =2 A= 2 \, \text{A}.

Find: Magnetic intensity HH at the center of the solenoid.

For a solenoid, the magnetic intensity is given by

H=nIH = nI

where nn is the number of turns per unit length.

First calculate turns per unit length:

n=12002=600 turns/mn = \frac{1200}{2} = 600 \, \text{turns/m}

Now substitute in the formula:

H=nI=600×2=1200 A m−1H = nI = 600 \times 2 = 1200 \, \text{A m}^{-1}

Thus, the magnetic intensity is 1.2×103 A m−11.2 \times 10^3 \, \text{A m}^{-1}. Therefore, the correct option is B.

Using turns per unit length explicitly

Given: A solenoid has 12001200 turns distributed uniformly over 2 m2 \, \text{m} and carries current 2 A2 \, \text{A}.

Find: The magnetic intensity HH.

The relevant relation is that inside a long solenoid,

H=nIH = nI

with

n=NLn = \frac{N}{L}

Substituting N=1200N = 1200 and L=2 mL = 2 \, \text{m},

n=12002=600 m−1n = \frac{1200}{2} = 600 \, \text{m}^{-1}

Then

H=600×2=1200 A m−1H = 600 \times 2 = 1200 \, \text{A m}^{-1}

So the required magnetic intensity is 1.2×103 A m−11.2 \times 10^3 \, \text{A m}^{-1}.

Common mistakes

  • Using the formula for magnetic field B=μ0nIB = \mu_0 nI instead of magnetic intensity H=nIH = nI. This is wrong because the question asks for magnetic intensity, not magnetic flux density. Use HH directly in units of A m−1\text{A m}^{-1}.

  • Taking n=1200n = 1200 instead of turns per unit length. This is wrong because nn means number of turns per metre, not total turns. First compute n=NLn = \frac{N}{L}.

  • Including the diameter of the solenoid in the calculation. This is unnecessary here because for magnetic intensity inside an ideal long solenoid, HH depends on turns per unit length and current, not on the diameter.

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