Question

a circular coil of 200 turns and of radius 20cm carries a current of 5A. calculate the magnetic induction at a point along its axis at a distance three times the radius of the coil from its center​

Answer 1

Answer:

A circular coil of $200$ turns and of radius $20cm$ carries a current of $5A$, the magnetic induction at a point along its axis at a distance three times the radius of the coil from its center​ is $3.14*10^{-4}T$

Explanation:

The magnetic field due to a circular coil of radius $r$ at a distance $x$ from its center, when the coil carrying a current $I$ is given by the equation

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

Here given values are

$n=200$

$r=20cm=0.2m$

$I=5A$

also given that the magnetic induction at a point $x$ along its axis is three times the radius of the coil that is $x=3r$

substitute the value in the first equation we get

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

$B=\frac{\mu _0nI}{2}\frac{r^2}{\left(10r)^2\right)^{\frac{3}{2}}}$

$B=\frac{\mu _0nI}{2*10r}$$=\frac{4\pi *10^{-7}*200*5*100 }{2*10*20} =3.14*10^{-4} T$