Question

Current $I_1$ and $I_2$ flow in the wires as shown in the figure. The field is zero at a distance $x$ to the right of O. Then,

• A. $x=\left(\frac{I_1}{I_2}\right)a$
• B. $x=\left(\frac{I_2}{I_1}\right)a$
• C. $x=\left(\frac{I_1-I_2}{I_2+I_2}\right)a$
• D. $x=\left(\frac{I_1+I_2}{I_1-I_2}\right)$

Answer 1

The correct option is C $x=\left(\frac{I_1-I_2}{I_2+I_2}\right)a$

Let's take a point P from point O at $x$ distance where magnetic field is zero.

Magnetic filed due to element AB and element DE will be zero as point lie on the same axis as these elements.
Magnetic field due to element BC and DF will be in opposite direction and they must be equal so that net magnetic field at point P is zero.
$\therefore B_1=B_2$
We know that magnetic field due to semi infinite wire is
$B=\frac{{\mu }_{0}I}{4\pi r}$

$\therefore \frac{{\mu }_0{I}_1}{4\pi (a+x)}=\frac{{\mu }_0{I}_2}{4\pi (a-x)}$

$\frac{a-x}{a+x}=\frac{I_2}{I_1}$

$\therefore x=\frac{(I_1-I_2)}{(I_1+I_2)}a$