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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?

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?

Grade:12

3 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
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) 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

Gitanjali Rout
184 Points
3 years ago
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
Gitanjali Rout
184 Points
3 years ago
\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) isr`from axis of wire(wherer`Magnetic field at a distanceB_1=\dfrac{\mu_0i}{2\pi r}is greater than radius of wire) isrfrom the axis of the wire(whereragnetic field at a distance

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