Question

As shown in the figure, two infinitely long, identical wires are bent by $90^\circ $ and placed in such a way that the segments $LP$ and $QM$ are along $x$ -axis, while segments $PS$ and $QN$ are parallel to $y-$ axis. Let $OP=OQ=5cm,$ and the magnitude of the magnetic field at $O$ is $10^{- 5}T,$ and also the two wires carry equal currents (see figure). What will be the magnitude of current in each wire in $10^{- 1}A$ ?

Answer 1

The magnetic field due to wires $LP$ and $MQ$ will be zero at point $O$ .
Magnetic field at point $O$ due to vertical wires is given as,
$B = B _{S P}+ B _{O N}$
$\therefore \quad B_{ SP }=\frac{\mu_0 I}{4 \pi r}\left(\sin 0+\sin 90^{\circ}\right)$
$B _{ QN }=\frac{\mu_0 I }{4 \pi r }\left(\sin 0+\sin 90^{\circ}\right)$
$B =2 \times \frac{\mu_0 I }{4 \pi r }\left(\sin 0+\sin 90^{\circ}\right)$
$\Rightarrow 10^{-5}=\frac{2 \times 4 \pi \times 10^{-7}}{4 \pi} \times \frac{1}{0.05}$
$\Rightarrow I =2.5=25 \times 10^{-1} A$