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Physics Question 38 – JEE-MAIN 2026

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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 _______ μT. (μ04π=10−7 N/A2)

Use the right-hand thumb rule to determine the direction of the magnetic field produced by each wire at the midpoint.

Step 1: Determine the direction of magnetic fields and total field✦ Active

The two parallel wires carry current in opposite directions. Using the right-hand thumb rule, at the midpoint between them, the magnetic field produced by each wire will be in the same direction (e.g., both into the page or both out of the page, depending on the chosen current directions). Therefore, the magnitudes of the individual magnetic fields add up to give the total magnetic field.

Step 2: Calculate the magnetic field due to a single wire○ Expand

The distance from each wire to the midpoint is r=d2=8 cm2=4 cm=0.04 m. The current in each wire is I=30 A. The magnetic field due to one long straight wire is given by B=μ0I2πr. We are given μ04π=10−7 N/A2, which implies μ02π=2×10−7 N/A2.

B1=B2=(2×10−7T⋅mA)×30 A0.04 m=60×10−70.04 T=1500×10−7 T=1.5×10−4 T
💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before calculation.

Step 3: Calculate the total magnetic field and convert units○ Expand

Since the fields add up, the total magnetic field at the midpoint is Btotal=B1+B2.

Btotal=2×(1.5×10−4 T)=3.0×10−4 T

To convert the result to microteslas (μT), we multiply by 106 (since 1 T=106μT).

Btotal=3.0×10−4×106μT=300μT
💡 Teacher's Secret Hint

Pay attention to the requested units for the final answer.

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