In the figure, there are two charges +Q and -Q on the x-axis at x = +a and x = -

Question

In the figure, there are two charges +Q and -Q on the x-axis at x = +a and x = -a. The charges are connected by a conducting bar such that a current I = -dQ/dt flows along the bar, as shown. It can be shown that the electric field due to the two point charges at any point in the yz plane a distance r from the origin is: E vector = Qa/2pi epsilon_0(r^2 + a^2)^3/2 (a) Find the displacement current through a circular area of radius R. (b) Assuming the point P in the figure is located at y = R, find the magnitude of the magnetic field at the point.

Explanation / Answer

to find displacement current wee first will find displacement current density
displacement current density, Jd = epsilon*dE/dt
where, epsilon is permittivity of free space
E is electric fierld in the region

consider at distance r form the origin in the yz plane a circle of thickness dr
Jd = epsilon*dE/dt
E = Qa/2*pi*epsilon(r^2 + a^2)^3/2
so dE/dt = -aI/2*pi*epsilon(r^2 + a^2)^3/2 [ because dQ/dt = -I, given ]
Jd = -aI/2*pi*(r^2 + a^2)^3/2

current = Jd*2*pi*rdr = -aI*rdr/(r^2 + a^2)^3/2 = dI
integrate above expression from r = 0 to r = R to find current through an area of radius R
so, intgrate [-aI*rdr/(r^2 + a^2)^3/2 ] from r = 0 to r = R
let r^2 + a^2 = t
2rdr = dt
intgrate [-aI*t^(-3/2)dt/2* ] = [aI*t^(-1/2) ] = [aI*(r^2 + a^2)^(-1/2) ]
applying limits
[aI*(R^2 + a^2)^(-1/2) ] - [aI*(a^2)^(-1/2) ] = [aI/sqroot((R^2 + a^2)) ] - [I ]
a. so Id = I(a/sqroot(R^2 + a^2) - 1])
b. from amperes law
B at point P is given by
B*2*pi*R = Id*mu [ mu is permeability of free space]
B = mu*I(a/sqroot(R^2 + a^2) - 1])/2*pi*R

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