If the current flowing through the shell of t | vedclass

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

Question

(B) The magnetic energy density $U$ in a magnetic field is given by $U = \frac{B^2}{2\mu_0}$.For a cylindrical shell of radius $R$ carrying current $i

Vedclass · Accepted Answer

Vedclass · Answer

(B) The magnetic energy density $U$ in a magnetic field is given by $U = \frac{B^2}{2\mu_0}$.For a cylindrical shell of radius $R$ carrying current $i$,the magnetic field $B$ at a distance $r = 2R$ (outside the shell) is given by Ampere's law as $B = \frac{\mu_0 i}{2\pi r}$.Substituting $r = 2R$,we get $B = \frac{\mu_0 i}{2\pi (2R)} = \frac{\mu_0 i}{4\pi R}$.Now,substituting this value of $B$ into the energy density formula:$U = \frac{1}{2\mu_0} \left( \frac{\mu_0 i}{4\pi R} \right)^2 = \frac{1}{2\mu_0} \cdot \frac{\mu_0^2 i^2}{16\pi^2 R^2} = \frac{\mu_0 i^2}{32\pi^2 R^2}$.Thus,$U \propto i^2$.The graph of $U$ versus $i$ is a parabola passing through the origin,symmetric about the $U$-axis,and having an increasing slope. This corresponds to the graph shown in option $B$.

If the current flowing through the shell of the previous objective is equal to $i$,then the energy density at a point at a distance $2R$ from the axis of the shell varies according to which graph?

Similar Questions

Identify the correct statement.

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil,the magnetic field at the centre is

The magnetic induction at $O$ due to the whole length of the conductor is:

The unit of magnetic permeability is

$A$ long straight wire in the horizontal plane carries a current of $50 \; A$ in the north to south direction. Give the magnitude and direction of $B$ at a point $2.5 \; m$ east of the wire.

Vedclass Test Series

Exam Paper Generator

Online Exam Module