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 $x\times {10}^{-5}\mathrm{T}\text{.}$ The value of $x$ is ________. 

[Take permeability of free space as ${\mu }_0=4\pi \times {10}^{-7}$ SI units.]

To find the resultant magnetic field at point $P$, we use the Biot-Savart law for the magnetic field due to a long straight current-carrying wire:

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

where:

• $\mu_0 = 4\pi \times 10^{-7}$ T·m/A (permeability of free space),
• $I$ is the current in the wire,
• $d$ is the perpendicular distance from the wire to the point.

Step 1: Magnetic Field Due to Wire Y at Point P

Wire $Y$ carries a current of $I_Y = 4A$ downward, and point $P$ is at a perpendicular distance of $d_Y = 4$ cm = 0.04 m. The magnetic field at $P$ due to $Y$ is:

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

 $$B_Y = \frac{16\pi \times 10^{-7}}{2\pi \times 0.04}$$

 $$B_Y = \frac{16 \times 10^{-7}}{0.08}$$ 

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

Using the right-hand rule, the direction of $B_Y$ at $P$ is out of the page.

Step 2: Magnetic Field Due to Wire X at Point P

Wire $X$ carries a current of $I_X = 5A$ upward, and the perpendicular distance from $X$ to $P$ is:

$$d_X = 6 + 4 = 10 \text{ cm} = 0.10 \text{ m}$$

The magnetic field at $P$ due to $X$ is:

$$B_X = \frac{(4\pi \times 10^{-7}) \times 5}{2\pi \times 0.10}$$

 $$B_X = \frac{20\pi \times 10^{-7}}{2\pi \times 0.10}$$ 

$$B_X = \frac{20 \times 10^{-7}}{0.20}$$ 

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

Using the right-hand rule, the direction of $B_X$ at $P$ is into the page.

Step 3: Resultant Magnetic Field at P

Since, fields $B_Y$ and $B_X$ are in the opposite direction , the net magnetic field (Out of the page) at $P$ is

$B_{net}=B_Y-B_X$

 $B_{\text{total}}=(2\times {10}^{-5})-(1\times {10}^{-5})$

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

Thus, the value of $x$ is 1.