An infinitely long conductor PQR is bent to f | vedclass

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Question

(C) The magnetic field at any point lying on the axis of a current-carrying straight conductor is zero.Case $1$: The current $I$ flows through $PQ$ an

Vedclass · Accepted Answer

$0.67$

Vedclass · Answer

(C) The magnetic field at any point lying on the axis of a current-carrying straight conductor is zero.Case $1$: The current $I$ flows through $PQ$ and $QR$. Point $M$ lies on the extension of $QR$. Therefore,the magnetic field at $M$ due to $QR$ is zero. The magnetic field $H_1$ at $M$ is only due to the segment $PQ$. Let $d$ be the perpendicular distance from $M$ to $PQ$. The magnetic field due to a semi-infinite wire is $H_1 = \frac{\mu_0 I}{4 \pi d}$.Case $2$: The current in $PQ$ is $I$,in $QR$ is $I/2$,and in $QS$ is $I/2$. Point $M$ lies on the extension of $QR$,so the magnetic field due to $QR$ is still zero. The magnetic field at $M$ is now $H_2 = H_{PQ} + H_{QS}$.Since the current in $PQ$ is unchanged,$H_{PQ} = H_1 = \frac{\mu_0 I}{4 \pi d}$.The conductor $QS$ is perpendicular to $PQ$ and $M$ lies on the perpendicular bisector of the line passing through $QS$ if we consider the geometry. However,$M$ is at a distance $d$ from $Q$ along the line $QR$. The magnetic field at $M$ due to a semi-infinite wire $QS$ carrying current $I/2$ is $H_{QS} = \frac{\mu_0 (I/2)}{4 \pi d} = \frac{1}{2} H_1$.Thus,$H_2 = H_1 + \frac{1}{2} H_1 = \frac{3}{2} H_1$.The ratio $H_1/H_2 = H_1 / (\frac{3}{2} H_1) = 2/3 \approx 0.67$.

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