Magnetic Force on Charges and Currents: Quick Revision

A brief note on Magnetic Force on Charges and Currents. Read it before a test. Try each question at the end before you open its answer.

The force on a moving charge

• Size: $F = qvB\sin\theta$, with $\theta$ the angle between $\vec{v}$ and $\vec{B}$. In vector form, $\vec{F} = q\,\vec{v} \times \vec{B}$.
• Direction: at right angles to $\vec{v}$ and $\vec{B}$. Right-hand rule for a positive charge. Reverse it for a negative charge.
• No force when the charge is at rest or moves along the field.
• Unit: $1\ \text{T} = 1\ \text{N}/(\text{A m})$. A gauss is $10^{-4}$ T.
• No work: the force is at right angles to the velocity. The speed does not change, only the direction.
• Lorentz force: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$.

Circles, helices, crossed fields

• Circle (velocity at right angles to the field): $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$.
• Time for a turn: $T = \dfrac{2\pi m}{qB}$. Frequency: $f = \dfrac{qB}{2\pi m}$. No speed in them.
• After a voltage $V$: $r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$.
• Helix (angle $\theta$ to the field): $r = \dfrac{mv\sin\theta}{qB}$. Pitch: $p = \dfrac{2\pi m\,v\cos\theta}{qB}$.
• Velocity selector: the forces cancel when $v = \dfrac{E}{B}$, for any charge and mass.
• Cyclotron: $f = \dfrac{qB}{2\pi m}$, and $K_{\max} = \dfrac{q^2B^2R^2}{2m}$. The energy does not depend on the gap voltage.

The force on a wire

• Straight wire: $F = BIL\sin\theta$. In vector form, $\vec{F} = I\,\vec{L} \times \vec{B}$, with $\vec{L}$ along the current.
• Direction: the right-hand rule, or Fleming's left-hand rule: forefinger field, middle finger current, thumb force.
• Bent wire in a uniform field: the force of the straight wire between its ends. A half circle gives $BI(2R)$. A closed loop gives zero.
• Floating wire: $BIL = mg$. Hung rod: $\tan\varphi = \dfrac{BIL}{mg}$.

A worked example

A proton enters a field of $0.50$ T at right angles, at $2.0 \times 10^{6}$ m/s. Find the radius and the time for a turn.

1. Radius. The top of $r = \dfrac{mv}{eB}$ is $mv = 1.67 \times 10^{-27}\ \text{kg} \times 2.0 \times 10^{6}\ \text{m/s} = 3.34 \times 10^{-21}\ \text{kg m/s}$. The bottom is $eB = 1.6 \times 10^{-19}\ \text{C} \times 0.50\ \text{T} = 8.0 \times 10^{-20}\ \text{C T}$. So $r = \dfrac{3.34 \times 10^{-21}\ \text{kg m/s}}{8.0 \times 10^{-20}\ \text{C T}} = 4.2 \times 10^{-2}\ \text{m}$.
2. Time for a turn. $T = \dfrac{2\pi m}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27}\ \text{kg}}{8.0 \times 10^{-20}\ \text{C T}} = 1.3 \times 10^{-7}\ \text{s}$.

Practise with new numbers

Check yourself

1. A wire is made twice as long, and the current in it is doubled. The field is the same and at right angles to the wire. By what factor does the force change?

2. A beam goes straight through crossed electric and magnetic fields. What does this tell you about its speed, and does the charge matter?

3. Why does a magnetic field not change the kinetic energy of a charge?