Topic 12.2 Magnetism and Moving Charges: apply the magnetic force on moving charges and currents, including circular motion and the force on a wire.

What this topic is asking

The College Board (Topic 12.2) wants you to apply the magnetic force on a moving charge, $\vec{F} = q\vec{v}\times\vec{B}$, use the right-hand rule, analyze the circular motion of a charge in a uniform field, and find the force on a current-carrying wire, $\vec{F} = I\vec{L}\times\vec{B}$.

The magnetic force on a moving charge

The cross product makes the force perpendicular to both the velocity and the field. Point the right hand's fingers along $\vec{v}$, curl toward $\vec{B}$; the thumb gives $\vec{v}\times\vec{B}$ (reverse it for a negative charge). A charge moving along the field feels no force.

No work, so circular motion

Because $\vec{F}\perp\vec{v}$ always, the magnetic force does no work: $\vec{F}\cdot\vec{v} = 0$. The speed and kinetic energy stay constant; only the direction turns. For a charge moving perpendicular to a uniform field, this perpendicular force is exactly centripetal, producing uniform circular motion. Setting the magnetic force equal to $\dfrac{mv^2}{r}$:

The period is independent of speed (the basis of the cyclotron). If the velocity also has a component along $\vec{B}$, the path becomes a helix.

Force on a current-carrying wire

A wire carrying current $I$ is a stream of moving charges, so it feels a magnetic force:

where $\vec{L}$ points along the wire in the current's direction and $\theta$ is the angle between the wire and the field. This is how motors generate torque.

Try this

Q1. A charge moves parallel to a magnetic field. State the magnetic force on it. [1 point]

• Cue. Zero: $F = qvB\sin\theta$ with $\theta = 0$.

Q2. An electron moves at $3.0\times10^6$ m/s perpendicular to a $0.20$ T field. Find the force magnitude. [2 points]

• Cue. $F = qvB = (1.6\times10^{-19})(3.0\times10^6)(0.20) = 9.6\times10^{-14}$ N.