Two long parallel wires X and Y, separated by a distance of 6 cm , carry currents of 5 A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is x × 10⁻⁵ T. The value of x is _________ . Take permeability of free space as μ₀=4 π × 10⁻⁷ SI units.

Solution

Physics sir · 4 steps

P is outside both wires, so measure 10 cm to X, and the two fields point opposite ways and subtract.

1. Distances to P. P lies 4 cm to the right of Y, on the side away from X. So P is 4 cm from Y and 6 + 4 = 10 cm from X.

   B = (μ₀ I)/(2π r) = (2 × 10⁻⁷)I/r

2. Each field. From X: B_X = 2 × 10⁻⁷ × 5/0.10 = 1 × 10⁻⁵ T. From Y: B_Y = 2 × 10⁻⁷ × 4/0.04 = 2 × 10⁻⁵ T.
3. Directions. By the right-hand rule, the upward current in X gives a field into the page at P. The downward current in Y gives a field out of the page at P. The fields oppose each other.
4. So. B_P = B_Y - B_X = 2 × 10⁻⁵ - 1 × 10⁻⁵ = 1 × 10⁻⁵ T, out of the page, so x = 1.

The common slip: 3. Adding the two fields as if they pointed the same way gives 3 × 10⁻⁵ T. At a point outside the pair, opposite currents give opposite fields, so they subtract.

The idea to keep. Opposite currents give opposite fields only at points outside the pair. Between them the fields add. Always fix each direction with the right-hand rule before you add or subtract.

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