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A current $I$ flows in a long thin walled cylinder(parallel to the axis) of radius $R$. What pressure do the walls experience?

This is the 263rd problem in Section III from the book 'Problems In General Physics by IE Irodov'.


My attempt:

Consider a thin segment of the cylinder parallel to the axis of thickness $Rd\theta$(i.e. subtends $d\theta$ at the center). As the current is uniform, the current in this segment will be $$dI = I\frac{d\theta}{2\pi}$$ The magnetic field $B$ near this segment will be $$B=\frac{\mu_0 I}{2\pi R}$$

Force on a length $l$ of this segment will be $$F=BIl=\frac{\mu_0 I}{2\pi R} I\frac{d\theta}{2\pi} l=\frac{\mu_0I^2ld\theta}{4\pi^2R}$$ Pressure will be $$P=\frac{F}{A}=\frac F{lRd\theta}=\frac{\mu_0I^2}{4\pi^2R^2}$$


The answer given in the book is $$\frac{\mu_0I^2}{8\pi^2R^2}$$

There are two problems after this also asking to find the pressure due to magnetic forces, in which my answer differs from the actual answer by a factor of $\frac 12$, just like above. Is there something I am missing in my understanding of pressure?

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1 Answer

up vote 1 down vote accepted

That is because the magnetic force acting on the infinitesimal surface is not $\mu _0 I^2/2\pi R=B$ but $0.5B$ . The factor of half is present because the field inside the cylinder is zero and outside it, is $B$ and therefore the "average value" is $B/2$ acting on the segment. Actually, a better and correct way of understanding this is by stating the field due to the infinitesimal segment as $B_1$ and the field due to the rest of the cylinder as $B_2$. Then outside the cylinder:-
$$B_1+B_2=B$$
and inside $$B_1 + B_2=0$$.
These equations tell you that $$|B_1|=|B_2|=|B|/2$$ and the $B_2$ component is along the net magnetic field and the $B_1$ component is opposite direction to the net magnetic field(inside the cylinder) and along it (outside the cylinder). The force on the infinitesimal segment is only the $B_2$ component and not $B_1$ component which is it's own field. And thus the factor of $1/2$ in your answer.

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