Two parallel wires of equal lengths are separated by a distance of $3 \mathrm{~m}$ from each other. The…
Two parallel wires of equal lengths are separated by a distance of $3 \mathrm{~m}$ from each other. The currents flowing through $1^{\text {st }}$ and $2^{\text {nd }}$ wire is $3 \mathrm{~A}$ and $4.5 \mathrm{~A}$ respectively in opposite directions. The resultant magnetic field at mid point between the wires $\left(\mu_0=\right.$ permeability of free space)
$\frac{\mu_0}{2 \pi}$
$\frac{3 \mu_0}{2 \pi}$
$\frac{7 \mu_0}{2 \pi}$
$\frac{5 \mu_0}{2 \pi}$
Solution
Using Biot Savart law,
$\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}}$
$\therefore \quad$ Magnetic field due to first wire:
$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}_1}{2 \pi \mathrm{r}}=\frac{\mu_0 \times 3}{2 \pi \times 1.5}=\frac{2 \mu_0}{2 \pi}$
$\therefore \quad$ Magnetic field due to second wire:
$\mathrm{B}_2=\frac{\mu_{\mathrm{o}} \mathrm{I}_2}{2 \pi \mathrm{r}}=\frac{\mu_{\mathrm{o}} \times 4.5}{2 \pi \times 1.5}=\frac{3 \mu_0}{2 \pi}$
$\therefore \quad$ Net field,
$\mathrm{B}=\mathrm{B}_1+\mathrm{B}_2=\frac{2 \mu_0}{2 \pi}+\frac{3 \mu_0}{2 \pi}=\frac{5 \mu_0}{2 \pi}$