The magnetic induction at the centre of a cur | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Magnetic field at any point on the axis of current carrying coil at a distance $x$ from the centre is given by :

$\mathrm{B}_{\mathrm{a}}=\frac{\mu

Vedclass · Accepted Answer

$20$

Vedclass · Answer

Magnetic field at any point on the axis of current carrying coil at a distance $x$ from the centre is given by :

$\mathrm{B}_{\mathrm{a}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}ight)^{3 / 2}}$

At the centre,

${\mathrm{x}=0}$

$	herefore $ ${\mathrm{B}_{\mathrm{e}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2 \mathrm{r}^{3}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}}}$

$	herefore $ ${\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}} 	imes \frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}ight)^{3 / 2}}{\mu_{0} \mathrm{nir}^{2}}}$

Given : $\quad \mathrm{B}_{\mathrm{c}}=5 \sqrt{5} \mathrm{B}_{\mathrm{a}}$

$	herefore $ $5 \sqrt{5}=\frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}ight)^{3 / 2}}{\mathrm{r}^{3}}=\frac{\left(100+\mathrm{x}^{2}ight)^{3 / 2}}{1000}$

Squaring on both sides.

${125=\frac{\left(100+x^{2}ight)^{3}}{10^{6}}} $

or  ${125 	imes 10^{6}=\left(100+x^{2}ight)^{3}}$

$100+x^{2}=5 	imes 10^{2}=500$

$	herefore \quad \mathrm{x}=20 \,\mathrm{cm}$

The magnetic induction at the centre of a current carrying circular coil of radius $10\, cm$ is $5\sqrt 5 \,times$ the magnetic induction at a point on its axis. The distance of the point from the centre of the coil (in $cm$) is

Similar Questions

Find the magnetic field at the point $P$ in figure. The curved portion is a semicircle connected to two long straight wires.

A closely wound flat circular coil of $25$ $turns$ of wire has diameter of $10\, cm$ and carries a current of $4\, ampere$. Determine the flux density at the centre of a coil

Find out magnetic field at point $O$ ?