Question

A circular coil of 100 turns and mean radius 0.04m carries a current of 3A. Calculate the magnetic field at a point 0.02m from its centre and on its axis.
$μo = 4\pi \times {10}^{ - 7}$
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Answer 1

Answer:

A circular coil of $100$ turns and the mean radius $0.04m$ carry a current of $3A$,the magnetic field at a point $0.02m$ from its center and on its axis is $110*10^{-5}T$

Explanation:

The magnetic field due to a circular coil is  at a distance $x$ from its center and on its axis is given by the formula,

$B=\mu _0nI\frac{r^2}{2\left(x^2+r^2\right)^{\frac{3}{2}}}$

where $n$- number of turns of the coil

$\mu _0=4\pi \times 10^{-7}$- permeability of free space

$x$- distance from the center of the coil

$r$- radius of the coil

$I$- current through the coil

From the question, it is given that

$n=100$

$x=0.02m$

$I=3A$

$r=0.04m$

substituting these values in the required equation we get,

$B=4\pi *10^{-7}*100*3(\frac{0.04^{2} }{2(x^{2}+r^{2} )^{3/2} } )$$=110.8*10^{-5}T$