Q15P A particle of charge聽q moves in... [FREE SOLUTION]

91影视

Introduction to Electrodynamics

David J. Griffiths

Physics

4th edition Edition 路 613 pages

ISBN: 9780321856562

Chapter 10: Q15P (page 455)

A particle of chargeq moves in a circle of radius a at constant angular velocity $蝇$ . (Assume that the circle lies in thexy plane, centered at the origin, and at time $t = 0$ the charge is at role="math" localid="1653885001176" $(a, 0)$ , on the positive x axis.) Find the Li茅nard-Wiechert potentials for points on the z-axis.

Short Answer

Expert verified

The Lienard-Wiechert potentials for points on the z-axis are $V (z, t) = \frac{1}{4 {蟺蔚}_{0}} \frac{q}{\sqrt{z^{2} + a^{2}}}$ and $A (z, t) = \frac{q 蝇 a}{4 {蟺蔚}_{0} c^{2} \sqrt{z^{2} + a^{2}}} [- \sin (蝇 t_{r}) \hat{x} + \cos (蝇 t_{r}) \hat{y}]$ .

Step by step solution

Expression for the position and linear velocity of a particle:

Write the expression for the position of a particle.

$r (t) = a [c o s (蝇 t) \hat{x} + s i n (蝇 t) \hat{y}]$

Here, t is the retarded time and $蝇 t$ is the position of q at time t.

Write the expression for the linear velocity of a particle.

$V (t) = a [- 蝇 \sin (蝇 t) \hat{x} + 蝇 \cos (蝇 t) \hat{y}] V (t) = 蝇 a [- \sin (蝇 t) \hat{x} + \cos (蝇 t) \hat{y}]$

Determine the Lienard-Wiechert potentials for a moving particle:

Write the expression for the retarded position to the field point r.

$r = z \hat{z} - a [\cos (蝇 t_{r}) \hat{x} + \sin (蝇 t_{r}) \hat{y}] r^{2} = (z \hat{z} - a [\cos (蝇 t_{r}) \hat{x} + \sin (蝇 t_{r}) \hat{y}]) (z \hat{z} - a [\cos (蝇 t_{r}) \hat{x} + \sin (蝇 t_{r}) \hat{y}]) r^{2} = z^{2} + a^{2} r = \sqrt{z^{2} + a^{2}}$

Write the relation between retarded position and linear velocity.

$\hat{r} 路 V = \frac{1}{r} (r 路 V)$

Substitute the value of rand Vin the above expression.

$\hat{r} 路 V = \frac{1}{r} [- a (- \sin 蝇 t_{r} \cos 蝇 t_{r} \hat{x} + \sin 蝇 t_{r} \cos 蝇 t_{r})] 路蝇 a [- \sin 蝇 t_{r} \hat{x} + \cos 蝇 t_{r} \hat{y}] \hat{r} 路 V = \frac{1}{r} [- 蝇 a^{2} (- \sin 蝇 t_{r} \cos 蝇 t_{r} + \sin 蝇 t_{r} \cos 蝇 t_{r})] \hat{r} 路 V = 0$

Write the expression for the Lienard-Wiechert potentials for a moving particle.

$V (r, t) = \frac{1}{4 {蟺蔚}_{0}} \frac{q c}{r c - \hat{r} 路 v}$ 鈥︹€� (1)

Here, v is the velocity of the charge, r is the vector from the retard position to the field point r, c is the speed of light and q is the charge.

For the z-axis, re-write the above expression.

$V (z, t) = \frac{1}{4 {蟺蔚}_{0}} \frac{q}{r} V (z, t) = \frac{1}{4 {蟺蔚}_{0}} \frac{q}{\sqrt{z^{2} + a^{2}}}$

Determine the Lienard-Wiechert potentials for a vector potential:

Write the expression for the vector potential.

$A (r, t) = \frac{渭_{0}}{4 蟺} \frac{q c v}{(r c - r 路 v)}$ 鈥︹€� (2)

From equations (1) and (2),

$A (r, t) = \frac{v}{c^{2}} V (r, t)$

For the z-axis, re-write the above expression.

$A (z, t) = \frac{v}{c^{2}} V (z, t)$

Substitute $v = 蝇 a (- \sin 蝇 t_{r}) \hat{x} + \cos (蝇 t_{r}) \hat{y}$ and $r = z \hat{z} - a [\cos (蝇 t_{r}) \hat{x} + \sin (蝇 t_{r}) \hat{y}]$ in the above expression.

$A (z, t) = \frac{蝇 a (- \sin 蝇 t_{r}) \hat{x} + \cos (蝇 t_{r}) \hat{y}}{c^{2}} [\frac{q}{4 {蟺蔚}_{0} \sqrt{z^{2} + a^{2}}}] A (z, t) = \frac{q 蝇 a}{4 {蟺蔚}_{0} c^{2} \sqrt{z^{2} + a^{2}}} [- \sin (蝇 t_{r}) \hat{x} + \cos (蝇 t_{r}) \hat{y}]$

Therefore, the Lienard-Wiechert potentials for points on the z-axis are equated as $V (z, t) = \frac{1}{4 {蟺蔚}_{0}} \frac{q}{\sqrt{z^{2} + a 2}}$ and $A (z, t) = \frac{q 蝇 a}{4 {蟺蔚}_{0} c^{2} \sqrt{z^{2} + a^{2}}} [- \sin (蝇 t_{r}) \hat{x} + \cos (蝇 t_{r}) \hat{y}]$ .