# A charged particle is traveling through a uniform magnetic field, with its velocity perpendicular to the field

## Question:

A charged particle is traveling through a uniform magnetic field, with its velocity perpendicular to the field direction. You learned that such a particle experiences a magnetic force that causes it to move in a circular path. Also, because it is moving, the charged particle creates its own magnetic field $$\vec{B}_{\mathrm{p}}$$.

(a) Derive an expression for the magnetic field magnitude $$B_{\mathrm{p}}$$ at the center of the circular path in terms of the magnitude $$B_{\mathrm{ext}}$$ of the external uniform magnetic field, the particle's orbit radius $$R$$, the charge $$q$$ on the particle, and the particle's mass $$m$$.

(b) Using $$c_{0}=1 / \sqrt{\mu_{0} \epsilon_{0}}$$, show that your expression can be cast in the form below, on

$\frac{q^{2}}{4 \pi \epsilon_{0} R} \frac{1}{m c_{0}^{2}} B_{\mathrm{ext}}$

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