Vanier 203-SN2-RE Electricity and Magnetism tutoring

A magnetic-force example: direction, then magnitude

Use this short example to practise the magnetic-fields part of the course. Consider a small charged particle in a uniform, steady magnetic field. We calculate only the magnetic force at the instant shown.

$$\begin{gathered}\vec F_B=q(\vec v\times\vec B)\\F_B=|q|vB\sin\theta\end{gathered}$$

Here q is the signed charge in coulombs (C); v is the speed in metres per second (m/s); B is the magnetic-field magnitude in teslas (T); θ is the angle between velocity and field; and F_B is the force magnitude in newtons (N). The cross product sets the direction; the magnitude uses |q|.

A particle with q = +2.0 nC moves right at 3.0 × 10⁴ m/s. A 0.40 T magnetic field points into the page. Find the magnetic-force magnitude and direction.

1. 1. Draw the directions

   Rightward velocity and an into-the-page field are perpendicular, so θ = 90°. Curl the fingers of your right hand from velocity toward field; your thumb points up. The charge is positive, so the magnetic force points up.

2. 2. Convert and calculate

   Convert 2.0 nC to 2.0 × 10⁻⁹ C. Then use the positive charge magnitude.

   $$\begin{aligned}F_B={}&(2.0\times10^{-9})(3.0\times10^4)\\&\times(0.40)\sin90^\circ\\={}&2.4\times10^{-5}\,\mathrm N\end{aligned}$$

3. 3. Check the result

   The force is 2.4 × 10⁻⁵ N upward. It is perpendicular to both velocity and field. The units agree: C × (m/s) × T = N. A magnetic force is perpendicular to the instantaneous velocity, so it does no work on the particle. With magnetic force alone, the direction of motion can change while the speed stays constant.

Try two checks