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(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Here,the distance between the wires

Vedclass · Accepted Answer

$\mu_0 / 2 \pi$

Vedclass · Answer

(C) The magnetic field $B$ due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.Here,the distance between the wires is $d = 5 \; m$. The point is at the midpoint,so the distance from each wire is $r = d/2 = 2.5 \; m$.For wire $P$ carrying $I_1 = 2.5 \; A$,the magnetic field $B_1 = \frac{\mu_0 I_1}{2 \pi r} = \frac{\mu_0 (2.5)}{2 \pi (2.5)} = \frac{\mu_0}{2 \pi}$.For wire $Q$ carrying $I_2 = 5 \; A$,the magnetic field $B_2 = \frac{\mu_0 I_2}{2 \pi r} = \frac{\mu_0 (5)}{2 \pi (2.5)} = \frac{2 \mu_0}{2 \pi} = \frac{\mu_0}{\pi}$.Since the currents are in the same direction,by the right-hand rule,the magnetic fields at the midpoint will be in opposite directions.The net magnetic field is $B_{net} = |B_2 - B_1| = |\frac{\mu_0}{\pi} - \frac{\mu_0}{2 \pi}| = \frac{\mu_0}{2 \pi}$.

Two long parallel wires $P$ and $Q$ are held perpendicular to the plane of the paper with a distance of $5 \; m$ between them. If $P$ and $Q$ carry currents of $2.5 \; A$ and $5 \; A$ respectively in the same direction,then the magnetic field at a point half-way between the wires is:

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