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

Answer: 25

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,
$B=B_{SP}+B_{ON}$
$\therefore B_{SP}=\frac{{\mu }_{0}I}{4\pi r}\left(\sin 0+\sin 90^{\circ }\right)$
$B_{QN}=\frac{{\mu }_{0}I}{4\pi r}\left(\sin 0+\sin 90^{\circ }\right)$
$B=2\times \frac{{\mu }_{0}I}{4\pi r}\left(\sin 0+\sin 90^{\circ }\right)$
$\Rightarrow 10^{-5}=\frac{2\times 4\pi \times 10^{-7}}{4\pi }\times \frac{1}{0.05}$
$\Rightarrow I=2.5=25\times 10^{-1}A$