Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point P midway between the wires is _______ μ T
(Given : μ₀=4 π × 10⁻⁷ TmA⁻¹)

Solution

Physics sir · 4 steps

Opposite currents, point between them: both fields point the same way, so just add them.

1. Distance to each wire. P is midway, so it is r = 2.5 cm = 0.025 m from each wire, not 5 cm.
2. Directions at P. Left wire, current up: by the right-hand rule its field to its right is into the page. Right wire, current down: its field to its left is also into the page. The two fields add.

   B_P = 2 × (μ₀ I)/(2π r) = 2 × (4π × 10⁻⁷)(10)/(2π (0.025))

3. Put in numbers. Each wire gives (2 × 10⁻⁷ × 10)/0.025 = 8 × 10⁻⁵ T = 80 μT.
4. So. B_P = 80 + 80 = 160 μT, so the answer is 160.

The common slip: 80. Using the full separation r = 5 cm instead of 2.5 cm gives 40 μT per wire and 80 in total, not 160.

The idea to keep. Between two parallel wires, opposite currents make the fields add and like currents make them cancel. Always measure r from each wire to the point, not the full separation.

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