JEE Main 2026 Apr 6 Shift 2, Physics Q13: Magnetic Field due to Electric Current

JEE Main2026Apr 6, Shift 2Physics

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 _____ $μ$ T. ( $\frac{μ _{0}}{4 π} = 1 0^{- 7}$ N/A $^{2}$ )

[ADD IMAGE: Two vertical long straight parallel wires separated by 8 cm. Left wire carries current I = 30 A upward; right wire carries current I = 30 A downward. Midpoint indicated between them.]

A
30
B
300
C
150
D
0.1

Solution

At the midpoint each wire is at distance $d = 4$ cm $= 4 \times 1 0^{- 2}$ m. For antiparallel currents, the fields at the midpoint add (same direction). Field due to one wire:

$B_{1} = \frac{μ _{0} I}{2 π d} = \frac{2 \times 1 0 ^{- 7} \times 30}{4 \times 1 0 ^{- 2}}$ T.

Total: $B = 2 B_{1} = \frac{4 \times 1 0 ^{- 7} \times 30}{4 \times 1 0 ^{- 2}} = 3 \times 1 0^{- 4}$ T $= 300 μ$ T.

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JEE Main 2026 Apr 6 Shift 2, Physics Q13: Magnetic Field due to Electric Current

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