Two long current carrying conductors are placed to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 μT. The equal current flowing in the two conductors is :

Solution

Physics sir · 4 steps

At the midpoint the two fields add only if the currents are opposite, so find the direction first, then I.

1. Pick the direction. The midpoint is the same distance from both wires, so each wire gives a field of the same size there. For currents in the same direction the two fields point opposite ways and cancel to zero. A non-zero field of 300 μT means the currents must be opposite, so the two fields add.
2. Field of one wire. A long straight wire gives B = (μ₀ I)/(2π d). Here d is the distance from each wire to the midpoint, d = 4 cm = 0.04 m, and μ₀/2π = 2 × 10⁻⁷ T m A⁻¹.

   Bₙₑₜ = 2 × (μ₀ I)/(2π d) = (2 × 2 × 10⁻⁷ × I)/0.04 = 10⁻⁵ I

3. Solve for I. Set 10⁻⁵ I = 300 × 10⁻⁶ T. This gives I = 30 A. Each wire gives 150 μT, and the two together give 300 μT.
4. So. The currents are 30 A in opposite directions, which is option (b).

The idea to keep. Midway between two equal currents, same direction gives zero field and opposite directions give double. Always measure d from each wire to the point, not between the wires.

Answer from the NTA final answer key. Solution by PrepWiser.