Two long parallel wires $X$ and $Y$, separated by a distance of $6\ \text{cm}$, carry currents of $5\ \text{A}$ and $4\ \text{A}$, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point $P$ at a distance of $4\ \text{cm}$ from wire $Y$ is $x\times10^{-5}\ \text{T}$. The value of $x$ is ______. Take permeability of free space as $\mu_0 = 4\pi\times10^{-7}$ SI units.
Numerical value type. Enter your answer.
Answer: 1
$P$ is outside the pair, $4\ \text{cm}$ from $Y$ and $10\ \text{cm}$ from $X$.
$$B_X = \frac{\mu_0}{2\pi}\cdot\frac{5}{0.10} = 2\times10^{-7}\times50 = 1\times10^{-5}\ \text{T}$$
$$B_Y = \frac{\mu_0}{2\pi}\cdot\frac{4}{0.04} = 2\times10^{-7}\times100 = 2\times10^{-5}\ \text{T}$$
The currents are opposite and $P$ is on the same side of both wires, so the two fields point in opposite directions:
$$B = 2\times10^{-5} - 1\times10^{-5} = 1\times10^{-5}\ \text{T} \Rightarrow x = 1$$
Solution by Sreeraj P, M.Sc Physics