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(A) Consider a small strip of width $dx$ on the surface of the cylinder. The current per unit length is $K = \frac{I}{2 \pi R}$.The magnetic field $B$

Vedclass · Accepted Answer

$\frac{\mu_0 I^2}{8 \pi^2 R^2}$

Vedclass · Answer

(A) Consider a small strip of width $dx$ on the surface of the cylinder. The current per unit length is $K = \frac{I}{2 \pi R}$.The magnetic field $B$ inside the cylinder is $0$,and outside is $0$. However,the magnetic field at the surface due to the current element is $B = \frac{\mu_0 K}{2} = \frac{\mu_0 I}{4 \pi R}$.The force per unit area (magnetic pressure) on a current-carrying surface is given by $P = \frac{1}{2} \mu_0 K^2$.Substituting $K = \frac{I}{2 \pi R}$:$P = \frac{1}{2} \mu_0 \left( \frac{I}{2 \pi R} \right)^2 = \frac{1}{2} \mu_0 \frac{I^2}{4 \pi^2 R^2} = \frac{\mu_0 I^2}{8 \pi^2 R^2}$.

$A$ current $I$ flows along the length of a thin-walled,long metallic hollow cylinder of radius $R$,distributed uniformly on its surface. The pressure on the wall will be:

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