Q7.38P - Introduction to Electrodynamics

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.

Step-by-Step Solution

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 1: Write the given data from the question.

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

Step 2: Determine the formula of the magnetic charge q m .

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.

Step 3: Determine the magnetic charge q m .

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

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

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