Question

A circular loop of radius $R$ carries current $I_2$ in a clockwise direction, as shown in the figure. The centre of the loop is at a distance $D$ above a long straight wire. What should be the magnitude and direction of the current $I_1$ in the straight wire so that the net magnetic field at the centre of the loop is zero?

• A. $\frac{\pi D}{R}{I}_2(\to )$
• B. $\frac{\pi D}{R}{I}_2(\leftarrow )$
• C. $\frac{\pi D}{4R}{I}_2(\to )$
• D. $\frac{\pi D}{4R}{I}_2(\leftarrow )$

Answer 1

The correct option is A $\frac{\pi D}{R}{I}_2(\to )$

The magnetic field at the centre of the loop due to the current in the loop is,

$B_2=\frac{{\mu }_o{I}_2}{2R}\text{Inwards}(\otimes )$

For the net field at the centre to be zero, the field ${\overrightarrow{B}}_1$, due to the straight conductor should be directed opposite to ${\overrightarrow{B}}_2$ i. e. normally outwards.

$B_1=\frac{{\mu }_o{I}_1}{2\pi D}\text{Outwards}(\odot )$

To produce this field, the direction of current $I_1$ must be $\text{towards right}$.

Thus, $B_1=B_2$

$\Rightarrow \frac{{\mu }_o{I}_1}{2\pi D}=\frac{{\mu }_o{I}_2}{2R}$

$\therefore I_1=\frac{\pi D}{R}{I}_2(\to )$

Hence, option $\left(a\right)$ is the correct answer.