Question

An infinitely long conductor $P Q R$ is bent to form a right angle as shown. A current $I$ flows through $P Q R$. The magnetic field due to this current at the point $M$ is $H_{1}$. Now, another infinitely long straight conductor $Q S$ is connected at $Q$, so that the current is $1 / 2$ in $Q R$ as well as in $Q S$. The current in $P Q$ remaining unchanged. The magnetic field at $M$ is now $H_{2}$. The ratio $H_{1} / H_{2}$ is given by

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{2}{3}$
• D. 2

Answer 1

Answer: (C)

Magnetic field at any point lying on the current-carrying straight conductor is zero.
Here, $H_{1}=$ magnetic field at $M$ due to current in $P Q$.
$H_{2}=$ magnetic field at $M$ due to $Q R+$ magnetic field at $M$ due to $Q S+$ magnetic field at $M$ due to $P Q$.
$=0+\frac{H_{1}}{2}+H_{1}=\frac{3}{2} H_{1} $
$\Rightarrow \frac{H_{1}}{H_{2}}=\frac{2}{3}$