Question

Question: Assuming that "Coulomb's law" for magnetic charges ( q_m) reads

$\mathbf{F}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{{\mathbf{qm}}_{\mathbf{1}}{\mathbf{qm}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}$

Work out the force law for a monopole moving with velocity through electric and magnetic fields E and B.

Short answer

Answer

The value of the magnetic charge is$\overrightarrow{F}=q_m\left(\overrightarrow{B}-{\mu }_0{\epsilon }_0\overrightarrow{\upsilon }\times \overrightarrow{E}\right)$
.

Step-by-step solution

Write the given data from the question.

Assuming that "Coulomb's law" for magnetic charges (q_m ) reads.

Determine the formula of the magnetic charge qm.

Write the formula of the magnetic charge.

$\overrightarrow{\mathbf{F}}={\mathbf{q}}_{\mathbf{m}}\frac{{\mu }_{\mathbf{0}}}{\mathbf{4}\pi }\frac{{\mathbf{Q}}_{\mathbf{m}}}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}$

Here, q_m is magnetic charge, Q_m is magnetic charge and r is radius.

Determine the magnetic charge qm.

Determine the magnetic charge q_m due to magnetic charge Q_m is given by:

$\overrightarrow{F}=q_m\frac{{\mu }_0}{4\pi }\frac{Q_m}{r^2}\hat{r}$

It is seen that magnetic field is given by $\overrightarrow{B}\equiv \frac{{\mu }_0}{4\pi }\frac{Q_m}{r^2}\hat{r}$. By analogy replace by $\overrightarrow{E}$ and $\overrightarrow{B}$by $-{\mu }_0{\epsilon }_0\overrightarrow{E}$(Griffiths page 339). Total force is given by

role="math" localid="1658304974224" $\overrightarrow{F}=q_m\left(\overrightarrow{B}-{\mu }_0{\epsilon }_0\overrightarrow{\upsilon }\times \overrightarrow{E}\right)$

Therefore, the value of the magnetic charge q_m is $\overrightarrow{F}=q_m\left(\overrightarrow{B}-{\mu }_0{\epsilon }_0\overrightarrow{\upsilon }\times \overrightarrow{E}\right)$.