An infinitely long conductor PQR is bent to f | vedclass

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(B) Let the distance from $Q$ to $M$ be $d$. The point $M$ lies on the line extending from $QR$. Therefore,the magnetic field at $M$ due to the segmen

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$1$

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(B) Let the distance from $Q$ to $M$ be $d$. The point $M$ lies on the line extending from $QR$. Therefore,the magnetic field at $M$ due to the segment $QR$ is zero.For the first case,the magnetic field $H_1$ at $M$ is due only to the segment $PQ$. Using the Biot-Savart Law for a semi-infinite wire,the magnetic field at a perpendicular distance $d$ from the end of the wire is $H_1 = \frac{\mu_0 I}{4 \pi d}$.In the second case,a conductor $QS$ is added. The current $I$ splits at $Q$. Since $PQ$ and $QS$ are collinear and $QR$ is perpendicular to them,the current $I$ splits into $I/2$ in $PQ$ and $I/2$ in $QS$ (assuming equal resistance per unit length). The segment $QR$ now carries no current.The magnetic field $H_2$ at $M$ is the sum of the fields due to $PQ$ and $QS$:$H_2 = H_{PQ} + H_{QS} + H_{QR}$$H_{PQ} = \frac{\mu_0 (I/2)}{4 \pi d} = \frac{1}{2} H_1$$H_{QS} = \frac{\mu_0 (I/2)}{4 \pi d} = \frac{1}{2} H_1$ (since $M$ is at distance $d$ from $Q$ along the line of $QS$)$H_{QR} = 0$Thus,$H_2 = \frac{1}{2} H_1 + \frac{1}{2} H_1 = H_1$.Therefore,the ratio $H_1/H_2 = 1$.

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