An infinitely long conductor is bent to form a right angle as shown in Figure. A current flows through . The magnetic field due to this current at the point is . Now, another infinitely long straight conductor is connected at so that current is in

Question

An infinitely long conductor $P Q R$ is bent to form a right angle as shown in Figure. 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 current is $I / 2$ 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 $\frac{H_{1}}{H_{2}}$ is given by [2000]

Smart Achievers · Accepted Answer

(3)

Magnetic field at $M$ due to $I$ and $Q R$

$H_{1} = \frac{1}{2} [\frac{μ_{0} I}{2 π R}] + 0 = \frac{μ_{0} I}{4 π R}$

[ $∵$ Magnetic field B = 0 at any point on the current-carrying straight conductor]

Now when wire $Q S$ is joined,

$H_{2}$ = (Magnetic field at M due to PQ) + (Magnetic field at M due to QR) + (Magnetic field at M due to QS)

$= \frac{1}{2} [\frac{μ_{0} I}{2 π R}] + 0 + \frac{1}{2} [\frac{μ_{0} I / 2}{2 π R}]$ $= \frac{3 μ_{0} I}{8 π R}$

$∴$ $\frac{H_{1}}{H_{2}} = \frac{\frac{μ_{0} I}{4 π R}}{\frac{3 μ_{0} I}{8 π R}} = \frac{2}{3}$

Solution

1 $\frac{1}{2}$

2 $1$

3 $\frac{2}{3}$

4 $2$

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