The current through ABC and A'B'C' is I. What | vedclass

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Question

(B) The magnetic field at point $P$ due to the semi-infinite wire $AB$ is given by $B_{AB} = \frac{\mu_0 I}{4\pi r}$ directed into the plane $(\otimes

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \left( \frac{2I}{r} \right)$

Vedclass · Answer

(B) The magnetic field at point $P$ due to the semi-infinite wire $AB$ is given by $B_{AB} = \frac{\mu_0 I}{4\pi r}$ directed into the plane $(\otimes)$.The magnetic field at point $P$ due to the segment $BC$ is $B_{BC} = 0$ because point $P$ lies on the axis of the wire $BC$.The magnetic field at point $P$ due to the semi-infinite wire $A'B'$ is $B_{A'B'} = \frac{\mu_0 I}{4\pi r}$ directed into the plane $(\otimes)$.The magnetic field at point $P$ due to the segment $B'C'$ is $B_{B'C'} = 0$ because point $P$ lies on the axis of the wire $B'C'$.Since all non-zero magnetic field contributions are directed into the plane, the net magnetic field at $P$ is the sum of these contributions:$B_{net} = B_{AB} + B_{A'B'} = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r} = \frac{2\mu_0 I}{4\pi r} = \frac{\mu_0}{4\pi} \left( \frac{2I}{r} \right)$.

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).

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