Question

Two parallel, long wires carry currents i₁ and i₂ with i₁ > i₂. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 µT. If the direction of i₂ is reversed, the field becomes 30 µT. The ratio i₁/i₂ is
(a) 4
(b) 3
(c) 2
(d) 1

Answer 1

(c) 2

The magnetic field due to the current-carrying long, straight wire at point a is given by
$B=\frac{{\mathrm{\mu }}_{\mathrm{o}}i}{2\mathrm{\pi }a}$
When both the wires carry currents i₁ and i_{2 ​}in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.
$B\text{'}=\frac{{\mathrm{\mu }}_{\mathrm{o}}{i}_1}{2\mathrm{\pi }a}-\frac{{\mathrm{\mu }}_{\mathrm{o}}{i}_2}{2\mathrm{\pi }a}=10\mathrm{\mu T}..\left(1\right)$
Here, a is the distance of the midpoint from both the wires.
When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.
$B\text{'}\text{'}=\frac{{\mathrm{\mu }}_{\mathrm{o}}{i}_1}{2\mathrm{\pi }a}+\frac{{\mathrm{\mu }}_{\mathrm{o}}{i}_2}{2\mathrm{\pi }a}=30\mathrm{\mu T}..\left(2\right)$
On solving eqs. (1) and (2), we get
$$\begin{gathered} i_1-i_2=10 \\ {i}_1+i_2=30 \\ \Rightarrow i_1=20,\mathit{}{i}_2=10 \\ \Rightarrow \frac{i_1}{i_2}=\frac{2}{1}=2 \end{gathered}$$