# GATE Papers >> EEE >> 2015 >> Question No 22

Question No. 22

A steady current I is flowing in the –x direction through each of two infinitely long wires at $y=±\frac{L}{2}$ as shown in the figure. The permeability of the medium is ${\mu }_{0}$. The $\stackrel{\to }{B}$ -field at (0,L,0) is

Solution of Question No 22 of GATE 2015 EEE Paper

The direction of curent in both the wires is in $-\widehat{\mathrm x}$ direction,

So, magnetic field at y = L due to both wires will be in $-\widehat{\mathrm z}$

direction as per Right hand thumb rule.

By Ampere's Circuital law,

For first wire:

$\oint {\overline{\mathrm{B}}}_{1}.\mathrm{d}\stackrel{\to }{\mathrm{I}}={\mathrm{\mu }}_{0}\mathrm{I}$

$\begin{array}{l}{\mathrm B}_1\times2\mathrm\pi\left(\frac{\mathrm L}2\right)={\mathrm\mu}_0\mathrm I\\{\mathrm B}_1=\frac{{\mathrm\mu}_0\mathrm I}{\mathrm{πL}}\end{array}$

For second wire,

since, the current in both wires is in same direction the magnetic field will be clockwise and will get added up at (0, L, 0)

$\begin{array}{l}\mathrm B\left(0,\mathrm L,0\right)={\mathrm B}_1+{\mathrm B}_2=\frac{{\mathrm\mu}_0\mathrm I}{\mathrm{πL}}\left[1+\frac13\right]\\\overrightarrow{\mathrm B}(0,\mathrm L,0)=-\frac{4{\mathrm\mu}_0\mathrm I}{3\mathrm{πL}}\widehat{\mathrm z}\end{array}$