# Cylinder with Uniform Current

For a uniform current that goes through a long cylinder (z-axis) please find

a) force exerted by the upper half on the lower half of the cylinder.

b) magnitude of this force for cylinder length of 1 meter, diameter of 1 mm and current of 10 A.

The current density inside the wire is

$j=I/S=I/(πa^2)$

Consider first the entire wire. Compute the magnetic field induction inside the wire at distance $r<a$. Take a circular path of radius $r<a$ and apply Ampere circuital law:

$∮ B(r)*d l=μ*I_{inside}$ or $∮ B(r)*d l=μ∯ j*ds$

$B(r)*2πr=μI/(πa^2 )*πr^2$ that is $B(r)=μI/(2πa^2 )*r$

If we cut the wire in half, since the current is uniformly distributed inside the wire, the magnetic field generated by one half at distance $r< a$ will be halved (the line and surface integrals above remain the same just now we have only half of total current):

$B(r)=μI/(4πa^2 )*r$

The element of volume on the lower half of the wire is $d V=1/2*2πrdr=πr*d r$ and the current through this volume element is

$d I(r)=j*d V=I/(πa^2 )*πr*d r=I r dr/a^2$

The magnetic force (per unit length) on this element of volume is

$d F/L=B(r)*d I(r)=μI/(4πa^2 )*r*I r dr/a^2$

So that the total force (per unit length) on the lower half is

$F/L=(μI^2)/(4πa^4 ) ∫_0^a r^2*d r=(μI^2)/(4πa^4 )*a^3/3=(μI^2)/12πa$

Since in both halves the currents have the same direction the force is attractive.

For a= 0.5 mm and I=10 A one has

$F/L=(4π*10^{-7}*100)/(12π*0.0005)=0.0067 N$

Since in both halves the currents have the same direction the force is attractive.