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Magnetism

## Motion of a charged particle in a uniform magnetic field

When the velocity of a charged particle $\vec v$ is perpendicular to a uniform $\vec B$, the particle moves around a circle in a plane perpendicular to $\vec B$.

There is always a centripetal force in a circular path, which in this case provided by magnetic force, therefore the radius of the circular path is
$\underbrace{qvB}_{F_B}=\frac{mv^{2}}{r} \quad \Rightarrow \quad r=\frac{mv}{qB}$

The time required to particle travel one circle or the period of motion is the circumference of the circle divided by the velocity of charged particle
$T=\frac{2πr}{v}=\frac{2πm}{qB}$

The angular speed of the particle $\omega$, which is called cyclotron frequency, is
$\omega=\frac v r =\frac {qB}m$

If the velocity of the charged particle makes some angle with uniform magnetic field, then it moves in a helical path.

If the particle’s velocity has components parallel and perpendicular to the uniform magnetic field $\vec B$, it moves in a helical path.

In the parallel case there is no force on the particle but in the perpendicular one there is centripetal acceleration toward the center.

The radius of the helical path and its period are
$r=\frac{mv_\bot}{qB}=\frac{m\left(v_0 \sin \theta \right)}{qB} \quad , \quad T=\frac{2πm}{qB}$