Magnetism> A solid cylindrical wire of radius R carr...

A solid cylindrical wire of radius R carries a current I the magnetic field is 5μT at a point which is 2R distance away from the axis of wire. Magnetic field at a point which is R/3 distance inside from the surface of wire is?

Syed suha , 7 Years ago

Grade 12

3 Answers

Eshan

Magnetic field at a distance $r$ from the axis of the wire(where $r$ is greater than radius of wire) is $B_1=\dfrac{\mu_0i}{2\pi r}$

Magnetic field at a distance $r'$ from axis of wire(where $r'$ is smaller than radius) is

$B_2=\dfrac{\mu_0i_{enc}}{2\pi r'}$ $=\dfrac{\mu_0i(\dfrac{r'^2}{R^2})}{2\pi r'}$ $=\dfrac{\mu_0 ir'}{2\pi R^2}$
$\implies \dfrac{B_2}{B_1}=\dfrac{r'r}{R^2}$
$\implies B_2=5\mu T\times \dfrac{(\dfrac{2R}{3})2R}{R^2}=6.66\mu T$

Last Activity: 7 Years ago

Gitanjali Rout

Magnetic field at a distancerfrom the axis of the wire(whereris greater than radius of wire) isB_1=\dfrac{\mu_0i}{2\pi r}Magnetic field at a distancer`from axis of wire(wherer`is smaller than radius) isB_2=\dfrac{\mu_0i_{enc}}{2\pi r`}=\dfrac{\mu_0i(\dfrac{r`^2}{R^2})}{2\pi r`}=\dfrac{\mu_0 ir`}{2\pi R^2}\implies \dfrac{B_2}{B_1}=\dfrac{r`r}{R^2}\implies B_2=5\mu T\times \dfrac{(\dfrac{2R}{3})2R}{R^2}=6.66\mu T

Last Activity: 7 Years ago

Gitanjali Rout

$\implies \dfrac{B_2}{B_1}=\dfrac{r`r}{R^2}$ $=\dfrac{\mu_0 ir`}{2\pi R^2}$ $=\dfrac{\mu_0i(\dfrac{r`^2}{R^2})}{2\pi r`}$ $B_2=\dfrac{\mu_0i_{enc}}{2\pi r`}$ is smaller than radius) is $r`$ from axis of wire(where $r`$ Magnetic field at a distance $B_1=\dfrac{\mu_0i}{2\pi r}$ is greater than radius of wire) is $r$ from the axis of the wire(where $r$ agnetic field at a distance

Last Activity: 7 Years ago