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(A, C) Using Ampere's law,the magnetic field $B$ at a distance $r$ from the axis (where $a < r < b$) is given by $B = \frac{\mu_0 I}{2\pi r}$.The ener

Vedclass · Accepted Answer

Self-inductance of the given arrangement per unit length is $\frac{\mu_0}{2\pi} \ln \left( \frac{b}{a} \right)$.

Vedclass · Answer

(A, C) Using Ampere's law,the magnetic field $B$ at a distance $r$ from the axis (where $a The energy density $u_B$ in the magnetic field is $u_B = \frac{B^2}{2\mu_0} = \frac{1}{2\mu_0} \left( \frac{\mu_0 I}{2\pi r} \right)^2 = \frac{\mu_0 I^2}{8\pi^2 r^2}$.The energy stored per unit length $U_l$ is obtained by integrating the energy density over the volume of a cylindrical shell of length $l=1$ and thickness $dr$:$U_l = \int_a^b u_B (2\pi r dr) = \int_a^b \frac{\mu_0 I^2}{8\pi^2 r^2} (2\pi r dr) = \frac{\mu_0 I^2}{4\pi} \int_a^b \frac{dr}{r} = \frac{\mu_0 I^2}{4\pi} \ln \left( \frac{b}{a} \right)$.Since the energy stored in an inductor is $U = \frac{1}{2} L I^2$,the self-inductance per unit length $L_l$ is given by $U_l = \frac{1}{2} L_l I^2$.Equating the two,$\frac{1}{2} L_l I^2 = \frac{\mu_0 I^2}{4\pi} \ln \left( \frac{b}{a} \right)$,which gives $L_l = \frac{\mu_0}{2\pi} \ln \left( \frac{b}{a} \right)$.Thus,both options $A$ and $C$ are correct.

$A$ long coaxial cable carries current $I$ along its surface. The current flows down the surface of the inner cylinder of radius $a$ and back along the outer cylinder of radius $b$.

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