JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 24)
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 $x \times 10^{-5} \mathrm{~T}$. The value of $x$ is _________ . Take permeability of free space as $\mu_0=4 \pi \times 10^{-7}$ SI units.
_22nd_January_Evening_Shift_en_24_1.png)
Answer
1
Explanation
_22nd_January_Evening_Shift_en_24_2.png)
$$\begin{aligned} & B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ & =-\frac{100 \mu_0}{4 \pi} \\ & =-100 \times 10^{-7} \\ & =-1 \times 10^{-5} \mathrm{~T} \end{aligned}$$
Comments (0)
