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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2026
  • Medium
  • MCQ

A current of \(30\) A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of \(8\) cm. The magnetic field at the mid point between the two wires is __________ \(\mu\)T. \(\left(\dfrac{\mu_0}{4\pi} = 10^{-7}\text{ N/A}^2\right)\)

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Verified Solution

Answer & Solution

Correct Answer

(B) \(300\)

Step-by-step Solution

Detailed explanation

Let the distance between the wires be \(d = 8\text{ cm}\). The distance of the midpoint from each wire is \(r = \dfrac{d}{2} = 4\text{ cm} = 0.04\text{ m}\).

The current in each wire is \(I = 30\text{ A}\).

Since the currents flow in opposite directions, the magnetic fields produced by both wires at the midpoint point in the same direction and add up.

The total magnetic field at the midpoint is:

\(B_{\text{net}} = B_1 + B_2 = \dfrac{\mu_0 I}{2\pi r} + \dfrac{\mu_0 I}{2\pi r} = \dfrac{\mu_0 I}{\pi r}\)

Substituting the given values:

\(B_{\text{net}} = \dfrac{4\pi \times 10^{-7} \times 30}{\pi \times 0.04}\)

\(B_{\text{net}} = \dfrac{120 \times 10^{-7}}{0.04}\)

\(B_{\text{net}} = 3000 \times 10^{-7}\text{ T}\)

\(B_{\text{net}} = 300 \times 10^{-6}\text{ T} = 300\ \mu\text{T}\)

Answer: \(300\)
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