Consider an infinitely long, straight cylindrical conductor of radius a carrying

Question

Consider an infinitely long, straight cylindrical conductor of radius a carrying a steady current. The current is distributed in such a way that the current density J 1S a function of r, the distance from the axis. Assume that J-à, Asin(kr), where A and k are constants. Find the magnetic flux density B inside and outside the conductor. Verify that the obtained magnetic fields satisfy the boundary condition for tangential components of the magnetic field at the boundary r a. Show that the magnetic field in the center of the wire is zero.

Explanation / Answer

consider infinitely long , straight cylinderical conductor of radius a, carrying steady current

current density J = A*sin(kr)
where r is radius of the cylinder, and current is along z axis

for r > a
from ampere's law
B*2*pi*r = mu*I
where mu is permeability of free space
dI = J*2*pi*r*dr = Asin(kr)*2*pi*r*dr
integrating from r = 0 to r = a
I = 2*pi*[(sin(ka) - ka*cos(ka))]/k^2
I = 2*pi(sin(ka) - ka*cos(ka))//k^2
hence

B = mu(sin(ka)- kacos(ka))/k^2*r

for 0 < r < a
I = 2*pi*[(sin(kr) - kr*cos(kr))]/k^2
hence
B = mu*[(sin(kr) - kr*cos(kr))]/k^2*r

at r = a
B1 = mu(sin(ka) - ka*cos(ka))/k^2*a
B2 = mu(sin(ka) - ka*cos(ka))]/k^2*a

B1 = B2

also, when r-> 0
taking limit
B = mu(k*cos(kr) - kcos(kr) + k^2*rsin(ka))/k^2
B = mu(rsin(ka)) = 0 for r = 0

hecne magnetic field is 0 at the center of the wire

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