Under Pressure

Consider a thin long current carrying wire of length $l$ on the cylinder having thickness $dx$ .

Now consider two other such wires placed symmetric to this wire . Analysing force of attraction due to these wires we get that that the net force will be inwards.

Let us say our thin wire and this new wire subtends an angle $\theta$ at the center.

Let us say that the thickness of other wires is also small such that they subtend angle $d\theta$ at the center.

We know that force of attraction $d(d(F)) = \frac{\mu_{0} i_{1} i_{2}}{2 \pi r} l = \frac{\mu_{0} I dx I d\theta}{16 {\pi}^3 R^2sin(\frac{\theta}{2})}$

We know that due to other two wires only radial forces would be added and tangential forces would get cancelled.

Hence Net force = $2F sin(\frac{\theta}{2})$ = $\frac{\mu_{0} I dx I d\theta}{8 {\pi}^3 R^2}$

Hence total force $dF = \int_{0}^{\pi} \frac{\mu_{0} I dx I d\theta}{8 {\pi}^3 R^2} = \frac{\mu_{0}}{4\pi} \frac{I^2 l dx}{2 \pi R^2}$

Hence , $P = \frac{dF}{dA} = \frac{dF}{l dx} = \frac{\mu_{0}}{4\pi} \frac{I^2}{2 \pi R^2} = \boxed{ {10}^{-5}}$

Firstly, we calculate the magnetic field at a point on the cylinder's wall. Consider a circle $C$ on the cylinder's wall, which lies on a plane perpendicular to the cylinder's axis. This circle is of radius $R$ and enclose a current of $1 \, A$ inside it. Then, by using Ampère's law, we obtain that $\displaystyle \oint_C \overrightarrow{B} \cdot \overrightarrow{\mathrm{d}\ell} = \mu_0 I$ . Since the current is uniformly distributed, we can rewrite the above equation as $\displaystyle B \cdot \oint_C \mathrm{d}\ell = \mu_0 I$ , or equivalently, $\displaystyle B \cdot 2\pi R = \mu_0 I \Leftrightarrow B = \frac{\mu_0 I}{2\pi R}$ . Now we apply the magnetic pressure formula $\displaystyle P_B = \frac{B^2}{2\mu_0}$ to obtain $\displaystyle P_B = \frac{\mu_0 I}{8 \pi^2 R^2} = 9.947 \times 10^{-6}$ .

Consider a disk of infinitely small thickness on the wall of the sphere. The current inside this disc is $I$ . Using Ampere's Law, we obtain: $\oint_{C} \vec{B} . \vec{dl} = \mu_0 I$ Here $\vec{dl}$ corresponds to a small length of the circumference of the disc, and the surface integral calculates this value over the disc.
By symmetry, we can conclude that $\vec{B}$ is uniform throughout the circumference, so we can take it out of the integral and obtain: $B \oint_{C} \vec{dl} = \mu_0 I$ But note that $\oint_{C} \vec{dl}$ is simply the perimeter of the disc, which is also equal to $2 \pi r$ . We substitute this value in the equation: $B 2 \pi r = \mu_0 I$ $\implies B= \frac{\mu_0 I}{2 \pi r}$ Let $P$ be the pressure this magnetic field exerts. Applying the Magnetic Pressure Formula , we obtain: $P= \frac{B^2}{2\mu_0}$ $\implies B= \frac{ \frac{\mu_0 I}{2 \pi r}}{2 \mu_0}$ $\implies P= \frac{\mu_0 I}{8 \pi^2 R^2}$ Substituting the values, we obtain $P= \boxed{9.947 \times 10^{-6} \ \text{Pascals} }$

consider a small element du where u is the angle . and length of conductor be L. the current through du will be I (2 pi)/du. now take another 2 elements du at an angle u/2 [both sides i.e right and left]from line joining initial angle du to center.we can consider all 3 elements as wire carrying current . now find resultant force on initial du by other. then dp= df/r*du L. intergrate it for u from 0 to pi

f[res.]=uo i^{2} L/16r[pi^{2}]cosu/2

dp= f[res.]/r duL

BY RIGHT HAND THUNB RULE dir. of force come inwards hence dir. od pressure is inwards

Consider the cylinder to be infinite number of diametrically opposite current carrying conductors. Now consider simply two such opposite conducting long wires. Let their lengths be l each. And let they carry each a small current of di amperes. They are separated by a distance 2R. Hence the magnetic fields at each of the wires are equal but opposite in direction (right hand screw rule). Let this field be B,

where B= (permeability of free space)/(2 pi 2R) [from ampere's circuital law]

Magnetic forces on the conductor will be equal and in opposite directions (or rather, radially inwards; use right hand thumb rule once again)

Therefore force on each = di l B

Net force by the two opposite conductors = 2Bldi

Integrating over the whole cylinder, net force is 2Bli.

Now area of curved surface = 2 pi Rl

Therefore, Pressure= Bi/(pi*R)

so Pressure=(1.2566x10^-6 x1)/(0.04x3.14159) = 995x10^-8 approx 10^-5

If someone looks closely (for an observer who sees 1A current), there will be an inward lorentz pressure caused "only" by the magnetic field by each of the "long current carrying wire" that "constitute" the cylinder. the electric stress is zero because the current carrying cylinder is electrically neutral in steady state, charges are moving but they are not accumulating, no net charge, no coulombic forces, thereby no electric stress.This is true If they were originally uncharged. (the fact that cylinder is uncharged is not mentioned in the question, I have taken that as "pre-understood"). This stress is due to the "magnetic fields" cast by the long wires which decays as (1/r). Electric stress in a uniformly charged cylinder was delivered by long charged wires who cast "electric field" which decayed as (1/r), therefore for the magnitude of stress I need not re-do the integral to find the pressure, I can reuse the result. Its an analogue of electrical stress, which was earlier given by (Sigma Squared, divided by 2 epsilon naught), Instead of sigma (surface charge density, we have kappa - linear current density given in the question as 100/8pi (Amperes per metre). Instead of 1/epsilon we have (U0) which is 4pi E-7. you plug that in and we get (1/2) (kappa squared* U0); 100*U0/(128pi^2) = 9.947E-6 Pa