Question

A charge q is moving slowly with a uniform velocity . Obtain an expression for the vector potential and the magnetic field at a distance r from it.

Answer 1

solution

we are considering charge is moving with uniform velocity V in direction +X axis.the charge is at origine at time t=0.at any instant t

its position are x=vt, y=z=0

then retarded time

t'=t - r'/c ................eq.1

where r' is distance to the point A from charge q at retarded time.

at the erlier time the charge was at position

x=vt'

therefore

$r'=\sqrt{(x-vt')^2+y^2+z^2}$ ........eq.2

using equation 1 and 2

$r'^2=c^2(t-t')^2=(x-vt')^2+y^2+z^2$

$(v^2-c^2)t^2-2(xv-c^2t)t'+x^2+y^2+z^2-c^2t^2=0$

this is the quadratic equation in t'

its solution can be found using shridhar aacharya formula.and

value of t' putting in equation 1

we get

r' = c(t - t') ........eq.3

scaler potential can be written for a charge q which is moving with velocity v is given by

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0(r'-v.r'/c)}$

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0(\sqrt{(x-vt)^2+(1-v^2/c^2)(y^2+z^2)}}$

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0\sqrt{(1-v^2/c^2)}(\sqrt{((x-vt)/(\sqrt(1-v^2/c^2))^2+(y^2+z^2)}}$

relation between vector potential and scaler potential is

$A=\frac{v\phi}{c^2}$

for a charge q at a time t and position (vt,0,0)

then vector potential is given by

$A_x=\frac{q}{4\pi\in_0\sqrt{(1-v^2)}(\sqrt{(x-vt)^2/((1-v^2)+(y^2+z^2)}}$ ...eq.4

and A_y=A_z=0 because we are considering that charge q is moving in +ve x direction.

now

megnetic field for charge q which is moving with unnifonrm velocity v is

$B=\nabla\times A$

putting the value of A_x , A_y , A_z then magnetic field will be

$\fcolorbox{red}{aqua}{$B=\frac{\mu_0 \space qv\times r}{4\pi c^2\space r^3}$}$

this magnetic field must be in z direction because Electric field E,manetic field B and velocity v are perpendicular to each other.