Question

Four very long, current carrying wires $1,2,3$ and $4$ in the same plane intersect to form a square $40cm$ on each side, as shown in figure. Find the magnitude of the current I in wire $2,$ so that the magnetic field at the centre of the square is zero. Assume that there are no cross-contacts between any pair of wires. Give your answer in ampere.

Answer 1

Answer: 2

${\vec{B}}_1+{\vec{B}}_2+{\vec{B}}_3+{\vec{B}}_{4\text{ at the centre }}=\overrightarrow{0}$
$\frac{{\mu }_0{I}_1}{2\pi a}\otimes +{\overrightarrow{B}}_2+\frac{{\mu }_0{I}_3}{2\pi a}\odot +\frac{{\mu }_0{I}_4}{2\pi a}\otimes =\overrightarrow{0}\dots \dots (i)$
Assume inward direction to be positive, then
on substituting $a=\frac{20}{100}m=0.2m$ in (1), we
get ${\vec{B}}_2=\frac{{\mu }_0}{2\pi a}{I}_3-I_1-I_4$
$\Rightarrow {\vec{B}}_2=\frac{{\mu }_0}{2\pi a}20-10-8=\frac{{\mu }_0}{2\pi a}2A$
${\vec{B}}_2$ is $\otimes$, so we conclude that the field due to wire $2$ must be directed inwards $\otimes .$ For the field due to wire $2$ to be directed inwards the current in the wire $2$ must be in the downward direction. Hence, we get,
from (1),
$\frac{{\mu }_{0}I}{2\pi a}=\frac{{\mu }_0}{2\pi a}2A$
$\Rightarrow I=2$ A downwards)