Magnetic Effects & Magnetism

JEE Main level, single correct, medium

16 of 276

Question 16 of 276

A current of 3030 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 88 cm. The magnetic field at the mid point between the two wires is __________ μ\muT. (μ04π=10−7 N/A2)\left(\dfrac{\mu_0}{4\pi} = 10^{-7}\text{ N/A}^2\right)

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Answer: (B) 300300

Let the distance between the wires be d=8 cmd = 8\text{ cm}. The distance of the midpoint from each wire is r=d2=4 cm=0.04 mr = \dfrac{d}{2} = 4\text{ cm} = 0.04\text{ m}.

The current in each wire is I=30 AI = 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:

Bnet=B1+B2=μ0I2πr+μ0I2πr=μ0IπrB_{\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:

Bnet=4π×10−7×30π×0.04B_{\text{net}} = \dfrac{4\pi \times 10^{-7} \times 30}{\pi \times 0.04}

Bnet=120×10−70.04B_{\text{net}} = \dfrac{120 \times 10^{-7}}{0.04}

Bnet=3000×10−7 TB_{\text{net}} = 3000 \times 10^{-7}\text{ T}

Bnet=300×10−6 T=300 μTB_{\text{net}} = 300 \times 10^{-6}\text{ T} = 300\ \mu\text{T}

Answer: 300300

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