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(B) For a wire of radius $a$ carrying a uniform current $I$,the magnetic field $B$ at a distance $r$ $(r < a)$ from the axis is given by Ampere's Law:

Vedclass · Accepted Answer

$U = \frac{\mu_0 I^2}{16\pi}$

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(B) For a wire of radius $a$ carrying a uniform current $I$,the magnetic field $B$ at a distance $r$ $(r Since the current is uniform,$I_{enclosed} = I \cdot (\frac{\pi r^2}{\pi a^2}) = I \frac{r^2}{a^2}$.Thus,$B(2\pi r) = \mu_0 I \frac{r^2}{a^2}$,which gives $B = \frac{\mu_0 I r}{2\pi a^2}$.The magnetic energy density $u_m$ is given by $u_m = \frac{B^2}{2\mu_0} = \frac{1}{2\mu_0} (\frac{\mu_0 I r}{2\pi a^2})^2 = \frac{\mu_0 I^2 r^2}{8\pi^2 a^4}$.The energy stored in a cylindrical shell of length $l$,radius $r$,and thickness $dr$ is $dU = u_m \cdot dV = u_m \cdot (2\pi r l dr)$.Integrating from $r=0$ to $r=a$: $U = \int_0^a \frac{\mu_0 I^2 r^2}{8\pi^2 a^4} (2\pi r l) dr = \frac{\mu_0 I^2 l}{4\pi a^4} \int_0^a r^3 dr = \frac{\mu_0 I^2 l}{4\pi a^4} [\frac{r^4}{4}]_0^a = \frac{\mu_0 I^2 l}{16\pi}$.Therefore,the energy stored per unit length is $\frac{U}{l} = \frac{\mu_0 I^2}{16\pi}$.

$A$ long straight wire of circular cross-section is made of a non-magnetic material. The wire is of radius $a$. The wire carries a current $I$ which is uniformly distributed over its cross-section. The energy stored per unit length in the magnetic field contained within the wire is

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