Two infinite length wires carry currents 8A a | vedclass

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(D) The magnetic field due to an infinite wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $1$ along the $X$-axis carrying cur

Vedclass · Accepted Answer

$\frac{5\mu_0}{\pi d}$

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(D) The magnetic field due to an infinite wire at a distance $d$ is given by $B = \frac{\mu_0 I}{2\pi d}$.For wire $1$ along the $X$-axis carrying current $I_1 = 8A$,the magnetic field at $P(0, 0, d)$ is $B_1 = \frac{\mu_0 (8)}{2\pi d} = \frac{4\mu_0}{\pi d}$.For wire $2$ along the $Y$-axis carrying current $I_2 = 6A$,the magnetic field at $P(0, 0, d)$ is $B_2 = \frac{\mu_0 (6)}{2\pi d} = \frac{3\mu_0}{\pi d}$.Since these two magnetic fields are mutually perpendicular,the net magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2}$.$B_{net} = \sqrt{\left(\frac{4\mu_0}{\pi d}\right)^2 + \left(\frac{3\mu_0}{\pi d}\right)^2} = \frac{\mu_0}{\pi d} \sqrt{16 + 9} = \frac{\mu_0}{\pi d} \sqrt{25} = \frac{5\mu_0}{\pi d}$.

Two infinite length wires carry currents $8A$ and $6A$ respectively and are placed along the $X$ and $Y$-axes. The magnetic field at a point $P(0, 0, d) \ m$ will be:

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