Question

$ABCD$ is a square in a $x-y$ plane as shown in the figure. three long, straight wire kept perpendicular to the plane, passing the corners $A,B$ and $C$ respectively. The wire at $A$ and $C$ carry equal current $I$,directed out of the plane $ABCD($along $+vez-$axis $)$. If the net magnetic field at $D$ is zero , then the current through wire at $B$ must be

• A. $\sqrt{2}I,along+z-axis$
• B. $2I,along-z-axis$
• C. $\sqrt{2}I,along-z-axis$
• D. $2I,along+z-axis$

Answer 1

The correct option is B $2I,along-z-axis$

Let $I^{{}^{\prime }}$ be the current through the straight wire at $B$.

Let the magnetic field at $D$ due to the current carrying wire through $A$ be $\overrightarrow{B_1}$

Let the magnetic field at $D$ due to the current carrying wire through $C$ be $\overrightarrow{B_2}$

$B_1=\frac{{\mu }_0{I}_1}{2\pi a}(-\hat{i})$ where $a$ is the side length of the square.

$B_2=\frac{{\mu }_0{I}_2}{2\pi BD}=\frac{{\mu }_0{I}_2}{2\pi a}(-\hat{j})$ where $a$ is the side length of the square

Given $I_1=I_2=I$

Net field at $C$ due to $I_1$ and $I_2$ is $\overrightarrow{B_1}+\overrightarrow{B_2}=-\frac{{\mu }_{0}I}{2\pi a}(\hat{i}+\hat{j})$

Since net field at $D$ is zero, the field at $D$ due to $B$ is $\frac{{\mu }_{0}I}{2\pi a}(\hat{i}+\hat{j})$

Distance between the current $I^{{}^{\prime }}$ and point $D$ is $\sqrt{2}a$

Field at $D$ due to current I' is $B_3=\frac{{\mu }_0{I}^{\prime }}{2\pi \sqrt{2}a}$

$\therefore \frac{{\mu }_0{I}^{\prime }}{2\pi \sqrt{2}a}=\left|\frac{{\mu }_{0}I}{2\pi a}\right(\hat{i}+\hat{j}\left)\right|=\frac{{\mu }_{0}I}{2\pi a}\left|\right(\hat{i}+\hat{j}|=\frac{{\mu }_{0}I}{2\pi a}\sqrt{2}$

$\Rightarrow I^{\prime }=2I$ along the $-vez-$axis.