- #1

Adgorn

- 130

- 18

- Homework Statement
- A relativistic particle with mass ##m## and positive charge ##q## moves in the ##[yz]## plane in a circle of radius ##R## under the influence of a uniform magnetic field ##B=B_0 \hat x##. Find the period of the motion.

- Relevant Equations
- ##\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0##

My solution was as follows:

$$\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0$$

The movement is in the ##[yz]## plane so ##|\overrightarrow v\times \overrightarrow B_0|=vB_0##, therefore: $$\biggr |\frac {dp} {dt}\biggr |= \frac {qvB_0} {c}.$$ On the other hand, $$p=\gamma m v=\gamma m R \omega$$ and since the force is perpendicular to the velocity, hence to the momentum: $$\biggr|\frac {dp} {dt}\biggr|=p \omega=\gamma mR \omega^2.$$ Comparing the expressions, I got $$ \frac {qR \omega B_0} {c}=\frac {mR \omega^2} {\sqrt {1-\frac {R^2 \omega^2} {c^2}}}$$

which after some algebra yielded $$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$ My examiner marked this as false without explaining why, but I can't find any fault with this solution, so is it correct?

$$\frac {d\overrightarrow p} {dt}=q \frac {\overrightarrow v} {c}\times \overrightarrow B_0$$

The movement is in the ##[yz]## plane so ##|\overrightarrow v\times \overrightarrow B_0|=vB_0##, therefore: $$\biggr |\frac {dp} {dt}\biggr |= \frac {qvB_0} {c}.$$ On the other hand, $$p=\gamma m v=\gamma m R \omega$$ and since the force is perpendicular to the velocity, hence to the momentum: $$\biggr|\frac {dp} {dt}\biggr|=p \omega=\gamma mR \omega^2.$$ Comparing the expressions, I got $$ \frac {qR \omega B_0} {c}=\frac {mR \omega^2} {\sqrt {1-\frac {R^2 \omega^2} {c^2}}}$$

which after some algebra yielded $$T=\frac {2\pi} {\omega}=2\pi \sqrt {\frac {m^2c^2} {q^2B_0^2}+\frac {R^2} {c^2}}.$$ My examiner marked this as false without explaining why, but I can't find any fault with this solution, so is it correct?