Question

Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, 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 \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is\( \_\_\_\_\_\) cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.) 

Hint:

When calculating the magnetic field due to two parallel currents, use the formula \( B = \frac{\mu_0 I}{2\pi r} \) for each wire. Add the magnetic fields vectorially if they are in opposite directions.

Answer 1

Correct Answer: 1

To find the distance \( x \) of point P from wire X, we need to consider the magnetic fields produced by both wires at point P. The magnetic field due to a straight current-carrying wire at a distance \( r \) is given by:

\( B = \frac{\mu_0 I}{2\pi r} \)

For wire Y, at distance 4 cm:

\( B_Y = \frac{\mu_0 \times 4}{2\pi \times 0.04} = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.04} = 10^{-5} \, \text{T} \)

For wire X, at distance \( x \):

\( B_X = \frac{\mu_0 \times 5}{2\pi x} = \frac{4\pi \times 10^{-7} \times 5}{2\pi x} = \frac{10^{-6}}{x} \, \text{T} \)

Since the currents are in opposite directions, their magnetic fields at point P are in opposite directions. Hence, the resultant magnetic field \( B \) is:

\( B = |B_X - B_Y| = 3 \times 10^{-5} \, \text{T} \)

Substituting values:

\( \left|\frac{10^{-6}}{x} - 10^{-5}\right| = 3 \times 10^{-5} \)

Consider \( \frac{10^{-6}}{x} > 10^{-5} \):

\( \frac{10^{-6}}{x} - 10^{-5} = 3 \times 10^{-5} \)

\( \frac{10^{-6}}{x} = 4 \times 10^{-5} \)

\( x = \frac{10^{-6}}{4 \times 10^{-5}} = \frac{1}{4} \times 10^1 = 2.5 \, \text{cm} \)