Question

A vertical conductor $X$ carries a downward current of $5A$. The flux density due to the current alone at a point $P10cm$ due east of $X$ is given by $x\times {10}^{-5}$. Find the value of '$x$'.

Answer 1

$I=5A,a=10cm=0.1m$

$B=\frac{{\mu }_{o}I}{2\pi a}=\frac{4\pi \times {10}^{-7}\times 5}{2\pi \times 0.1}=1\times {10}^{-5}T$

At P, the earth's horizontal magnetic flux density,

$B_e=4\times {10}^{-5}T$ (from South to North)

The direction of B is from north to south.

$\therefore$ Resultant intensity at $P=4\times {10}^{-5}-1\times {10}^{-5}T$ $=3\times {10}^{-5}T$ (From south to north)

For a point 10 cm, north of X the flux density due to the current in $X=1\times {10}^{-5}T$ (due east).

Therefore, $x$ = 1