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MCQMediumJEE 2025Biot–Savart Law

JEE Physics 2025 Question with Solution

An infinite wire has a circular bend of radius aa, and carrying a current II as shown in the figure. The magnitude of the magnetic field at the origin OO of the arc is given by:

A wire forms a three-fourth circular arc centered at O with radius a, then extends horizontally to the right and also as a lower straight segment to infinity, with current directions marked on the arc and straight parts.
  • A

    μ0I4πa(3π2+2)\frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} + 2 \right)

  • B

    μ0I2πa(π2+2)\frac{\mu_0 I}{2 \pi a} \left( \frac{\pi}{2} + 2 \right)

  • C

    μ0I4πa(3π2)\frac{\mu_0 I}{4 \pi a} \left( \frac{3\pi}{2} \right)

  • D

    μ0I2πa(3π2+1)\frac{\mu_0 I}{2 \pi a} \left( \frac{3\pi}{2} + 1 \right)

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: An infinite wire has a circular bend of radius aa carrying current II. The field is required at the origin OO.

Find: The magnitude of the magnetic field at OO.

Use superposition and add the magnetic field contributions of the different parts of the wire.

For the circular arc of angle 3π2\frac{3\pi}{2}, the magnetic field at the center is

Barc=μ0Iθ4πaB_{\text{arc}} = \frac{\mu_0 I \theta}{4\pi a}

with

θ=3π2\theta = \frac{3\pi}{2}

Therefore,

Barc=μ0I4πa(3π2)B_{\text{arc}} = \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2}\right)

Extracted Solution and Discrepancy Note

Given: the solution states: for the arc with radius aa and angle 3π2\frac{3\pi}{2}, the field at the origin is written, and then the total field is concluded as

B=μ0I4πa(3π2)B = \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2}\right)

Find: The correct option.

The solution explicitly marks The Correct Option is C and concludes

B=μ0I4πa(3π2)B = \frac{\mu_0 I}{4\pi a}\left(\frac{3\pi}{2}\right)

Hence, the correct option is C.

There is an internal inconsistency in the extracted working text because it labels the arc and straight-segment contributions incorrectly, but the final conclusion and the declared correct option both match option C.

Common mistakes

  • Students often use the formula for a full circular loop, which is wrong because the bend is only an arc. Use the arc-field formula with the actual angle θ=3π2\theta = \frac{3\pi}{2} instead.

  • A common error is to add magnetic field contributions from straight segments without checking geometry. First decide whether the origin lies on the line of the segment or whether the segment contributes nonzero field at OO.

  • Some students ignore the direction of current around the arc and add magnitudes blindly. Use the right-hand rule for each segment before combining contributions.

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