Two long parallel wires X and Y, separated by a distance of 6 cm , carry currents of 5A 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 $$x \times 10^{-5}T$$. The value of is__________ . Take permeability of free space as

$$\mu_{\circ}=4\pi \times 10^{-7}$$SI units.

$$I_X = 5 \text{ A}$$

$$I_Y = 4 \text{ A}$$

Distance between wires X and Y is $$6 \text{ cm}$$

Point P is at a distance of $$4 \text{ cm}$$ to the right of wire Y.

The total distance of point P from wire X is:

$$ r_X = 6 \text{ cm} + 4 \text{ cm} = 10 \text{ cm} = 0.1 \text{ m} $$

The distance of point P from wire Y is:

$$ r_Y = 4 \text{ cm} = 0.04 \text{ m} $$

The magnetic field due to a long straight wire is given by the formula:

$$ B = \frac{\mu_0 I}{2\pi r} $$

Calculate the magnetic field at point P due to wire X ($$B_X$$):

$$ B_X = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1} $$

$$ B_X = \frac{2 \times 10^{-7} \times 5}{0.1} $$

$$ B_X = 10 \times 10^{-6} \text{ T} = 1 \times 10^{-5} \text{ T} $$

By the right-hand grip rule, with current flowing upwards, the direction of $$B_X$$ at point P is into the plane of the paper.

Calculate the magnetic field at point P due to wire Y ($$B_Y$$):

$$ B_Y = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.04} $$

$$ B_Y = \frac{2 \times 10^{-7} \times 4}{0.04} $$

$$ B_Y = 2 \times 10^{-5} \text{ T} $$

By the right-hand grip rule, with current flowing downwards, the direction of $$B_Y$$ at point P is out of the plane of the paper.

Since the magnetic fields are in opposite directions, the resultant magnetic field ($$B_{net}$$) is the difference between their magnitudes:

$$ B_{net} = B_Y - B_X $$

$$ B_{net} = 2 \times 10^{-5} \text{ T} - 1 \times 10^{-5} \text{ T} $$

$$ B_{net} = 1 \times 10^{-5} \text{ T} $$

Comparing this with the given expression $$x \times 10^{-5} \text{ T}$$, we get the value of $$x$$:

$$ x = 1 $$