Question

Four very long, current carrying wires $1,2,3$ and $4$ in the same plane intersect to form a square $40\,cm$ 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

$\vec{B}_{1}+\vec{B}_{2}+\vec{B}_{3}+\vec{B}_{4 \text { at the centre }}=\overrightarrow{0}$
$\frac{\mu_{0} I _{1}}{2 \pi a } \otimes+\overrightarrow{ B }_{2}+\frac{\mu_{0} I _{3}}{2 \pi a } \odot+\frac{\mu_{0} I _{4}}{2 \pi a } \otimes=\overrightarrow{0} \ldots \ldots(i)$
Assume inward direction to be positive, then
on substituting $a=\frac{20}{100} m=0.2 \, m$ in (1), we
get $\vec{B}_{2}=\frac{\mu_{0}}{2 \pi a} I_{3}-I_{1}-I_{4}$
$\Rightarrow \vec{B}_{2}=\frac{\mu_{0}}{2 \pi a} 20-10-8=\frac{\mu_{0}}{2 \pi a} 2 A$
$\vec{B}_{2}$ 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),
$\frac{\mu_{0} I}{2 \pi a}=\frac{\mu_{0}}{2 \pi a} 2 A$
$\Rightarrow I=2$ A downwards)