Quick questions on Charges in magnetic fields explained: HSC Physics Module 6

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force:

Because $\vec{F} \perp \vec{v}$, the dot product $\vec{F} \cdot d\vec{s} = \vec{F} \cdot \vec{v} \, dt$ is always zero. The work done over any displacement is zero, so the kinetic energy and hence the speed are constant. The magnetic force can change a particle's direction but never its speed.

When a charged particle moves perpendicular to a uniform magnetic field, the constant-magnitude force perpendicular to $\vec{v}$ provides exactly the centripetal force needed for circular motion. Setting magnetic = centripetal:

1. Point the fingers of your right hand in the direction of $\vec{v}$. 2. Curl them toward $\vec{B}$ (through the smaller angle).

A singly-ionised carbon-12 atom is accelerated from rest through 500 V, then enters a uniform magnetic field of $0.20$ T perpendicular to its velocity. Find the radius of its circular path. ($m = 1.99 \times 10^{-26}$ kg, $q = 1.60 \times 10^{-19}$ C.)

$F = qvB$ is only the magnitude when $\vec{v} \perp \vec{B}$. In other cases use $F = qvB \sin \theta$.

$F = qE$ is the electric-field force. The magnetic-field force is $F = qvB \sin \theta$. They look similar; mix them up and you lose easy marks.

Always note whether the charge is positive or negative and flip the direction for an electron.

It changes the direction only. Speed and kinetic energy are constant.

The period depends only on $m/(qB)$. Faster particles orbit on larger circles in the same time, not faster.

$r = mv/(qB)$, not $r = qB/(mv)$ or $r = mvB/q$.