Question

The magnetic field at the centre of a current carrying circular coil of radius 10 cm is 5.5 times the magnetic
field at a point on its axis. The distance of the point from the centre of the coil (in cm) is

Answer 1

The magnetic field at the center of a current  carrying circular loop of radius R is: Bcenter=2Rμ0i

The magnetic field on the axis  at a distance  x from the center  of a current carrying circular loop of radius R is: Baxis=2(R2+x2)3/2μ0iR2

Given Bcenter=55Baxis

∴2Rμ0i=552(R2+x2)3/2μ0iR2

⇒R6(R2+x2)3=(55)2=125

⇒(R2R2+x2)3=125

Answer 2

Answer:

The magnetic field at the center of a current-carrying circular coil of radius  $10cm$ is $5.5$ times the magnetic  field at a point on its axis. The distance of the point from the center of the coil is $8.1cm$

Explanation:

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

$B=\frac{\mu _0I}{2}\frac{r^2}{\left(r^2+x^2\right)^{\frac{3}{2}}}$

At the center of the coil the distance $x=0$ so the magnetic field at the center is

$B_{c} =\frac{\mu _0I}{2}\frac{r^2}{\left(r^2\right)^{\frac{3}{2}}}=\frac{\mu _0I}{2r}$

The magnetic field at a distance $x$ is $5.5$ times the field at the center, that is

$B_{x}=5.5 \frac{\mu _0I}{2r}$

To find the distance on its axis we need to equate the first equation with the last one

$B=\frac{\mu _0I}{2}\frac{r^2}{\left(r^2+x^2\right)^{\frac{3}{2}}}$ $=5.5 \frac{\mu _0I}{2r}$

similar terms will cancel out and we get

$\frac{r^3}{\left(r^2+x^2\right)^{\frac{3}{2}}}=5.5$

take the power of $2/3$ on each side of the equation

$\frac{r^2}{\left(r^2+x^2\right)}=3\\2r^{2} =3x^{2} \\x=\sqrt{\frac{2}{3} }*r =0.81*10=8.1cm$