# 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

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

## Step by Step Answer:

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