Question

“Derive” the Lorentz force law, as follows: Let chargeqbe at rest in$\overline{\mathbf{S}}$, so $\overline{\mathbf{F}}\mathbf{=}\mathbf{q}\overline{\mathbf{E}}$, and let $\overline{\mathbf{S}}$move with velocity$\mathbf{v}\mathbf{=}\stackrel{\mathbf{\wedge }}{\mathbf{vx}}$with respect to S. Use the transformation rules (Eqs. 12.67 and 12.109) to rewrite $\overline{\mathbf{F}}$in terms of F, and $\overline{\mathbf{E}}$in terms of E and B. From these, deduce the formula for F in terms of E and B.

Short answer

The Lorentz force is deduced as$F=qE+q(v\times B)$

Step-by-step solution

Expression for Maxwell’s equation:

Using equation 12.67, write the equation for the transformation of forces from one frame to another frame.

${\overline{F}}_1=\frac{1}{y}{F}_{\perp }{\overline{F}}_1=F_1$

Here, $\mathrm{y}$ is the constant pertains to the relative motion between the two frames.

Deduce the Lorentz force law:

Write the expression for the force acting on the charge in the frame $\overline{S}$.

$\overline{F}=q\overline{E}$

Here, q is the charge and $\overline{E}$is the electric field.

Write the above expression in a vector form.

$\overline{F}=q{\overline{E}}_x\hat{x}+q{\overline{E}}_y\hat{y}+q{\overline{E}}_z\hat{z}$

Here, ${\overline{E}}_x,{\overline{E}}_y$and ${\overline{E}}_z$are the components of an electric field in the frame .

Write the expression for the force acting on the charge in frame S.

$F=F_x\hat{x}+F_y\hat{y}+F_z\hat{z}$ …… (1)

Here, $F_x,F_y$and $F_z$are the components of the forces in frame S.

Write the equations for the component of the forces of frame S in terms of the component of the forces of the frame .

$$\begin{gathered} F_x=q{\overline{E}}_x \\ {F}_y=\frac{1}{\gamma }q{\overline{E}}_y \\ {F}_z=\frac{1}{\gamma }q{\overline{E}}_z \end{gathered}$$

Using equation 12.109, the above component of the forces becomes,

$$\begin{gathered} F_x=q{E}_x \\ {F}_y=\frac{1}{\gamma }q\left(\gamma \right(E_y-v{B}_z\left)\right)=q(E_y-v{B}_z) \\ {F}_z=\frac{1}{\gamma }q\left(\gamma \right(E_z+v{B}_z\left)\right)=q(E_z-v{B}_y) \end{gathered}$$

Substitute $q{E}_x$for $F_x,q(E_y-v{B}_z)$for $F_y$and $q(E_z+v{B}_y)F_z$for $F_z$in equation (1).

$$\begin{gathered} F=q{E}_x\hat{x}+q(E_y-v{B}_z)\hat{y}+q(E_z+v{B}_y)\hat{z} \\ F=q\left[(E_x\hat{x}+E_y\hat{y}+E_z\hat{z})\right]-q\left[\left(v{B}_z\right)\hat{y}+\left(v{B}_y\right)\hat{z}\right].......\left(2\right) \end{gathered}$$

Here, $q\left[\left(v{B}_z\right)\hat{y}+\left(v{B}_y\right)\right]\hat{z}$and $q(E_x\hat{x}+E_y\hat{y}+E_z\hat{z})=qE$.

Hence, the equation (2) becomes,

$F=qE+q(v\times B)$

Therefore, the Lorentz force law is deduced.