Question

Two parallel long wires carry currents $i_1$ and $i_2$ with $i_1>i_2$. when the currents are in the same direction, the magnetic field at a point midway between the wires is $10\mu \text{T}$. If the direction of $i_2$ is reversed, the field becomes $30\mu \text{T}$. The ratio $\frac{i_1}{i_2}$ is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Initially when wires carry currents in the same direction, the magnetic field at mid point $\text{O}$ due to wire $1$ and $2$ respectively are,


$B_1=\frac{{\mu }_0{i}_1}{2\pi x}$ (Inwards to the plane of two wires)

$B_2=\frac{{\mu }_0{i}_2}{2\pi x}$ (Outward to the plane of wire)

$\because i_1>i_2$ hence $B_1>B_2$ and the direction of $B_1\&B_2$ are opposite

$\Rightarrow B_{net}=B_1-B_2$

or, $B_{net}=\frac{{\mu }_0}{2\pi x}(i_1-i_2)$

Also, given in question;

$\frac{{\mu }_0}{2\pi x}(i_1-i_2)=10\times {10}^{-6}$ ............ $\left(1\right)$

If the direction of $i_2$ is reversed then,


$B_1=\frac{{\mu }_{0}i}{2\pi x}$

$B_2=\frac{{\mu }_0{i}_2}{2\pi x}$

Both the fields are perpendicularly inward to the plane of wires.

$\Rightarrow B_{net}=B_1+B_2$

or, $B_{net}=\frac{{\mu }_0}{2\pi x}(i_1+i_2)$

According to question, in this case

$B_{net}=30\times {10}^{-6}\text{T}$

$\frac{{\mu }_0(i_1+i_2)}{2\pi x}=30\times {10}^{-6}....\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\frac{i_1+i_2}{i_1-i_2}=\frac{30\times {10}^{-6}}{10\times {10}^{-6}}$

or, $\frac{i_1+i_2}{i_1-i_2}=\frac{3}{1}$

or, $3i_1-3i_2=i_1+i_2$

or, $2i_1=4i_2$

$\therefore \frac{i_1}{i_2}=\frac{2}{1}$

Why this Question ? Note: The magnetic field due to a long current carrying wire at a perpendicular distance $x$ is given by, $B=\frac{{\mu }_{0}i}{2\pi x}$ In order to find the net field due to two or more wires, apply superposition principle by carefully taking into account, the direction of magnetic field due to different wires