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Question No. 22 EEE | GATE 2015

A steady current I is flowing in the –x direction through each of two infinitely long wires at y=±L2 as shown in the figure. The permeability of the medium is μ0. The B -field at (0,L,0) is


Answer : (A)  - 4 μ0 I3π LZ^


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:

B¯1.dI=μ0I

$ \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,

B¯2.dl=μ0IB2×2π3L2=μ0I, B2=μ0I3πL

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} $

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