Question

What is the force exerted on a moving charge of –2.0 μC at a 20° angle through a magnetic field of 3.0 × 10–4 T with a velocity of 5.0 × 106 m/s?

3.0 × 10–3 N
1.0 × 10–3 N
1.0 × 103 N

Can someone please give a full-length explanation of how to solve this

Answer 1

Approximately
`1.0 * 10^(-3)\; {\rm N}`.

Step-by-step explanation:

Consider a charge of magnitude
`q`. Assume that this charge is moving at a speed of
`v` across a magnetic field of strength
`B`. Let the angle between the magnetic field and the motion of the charge be
`\theta`.

Apply the following equation to find the magnitude of the magnetic force on this charge:

`F = q\, v\, B\, \sin(\theta)`.

Ensure that all quantities (
`q`,
`v`, and
`B`) are in standard units:

`q = 2.0\; {\rm \mu C} = 2.0 * 10^(-6)\; {\rm C}`.

Note that the unit of magnetic field strength, Tesla, is equivalent to
`{\rm N \cdot s \cdot C^(-1)\cdot m^(-1)}`.

The magnitude of the magnetic force on this moving charge would be:

`\begin{aligned}F &= q\, v\, B\, \sin(\theta)\\ &= (2.0 * 10^(-6)\; {\rm C}) \\ &\quad\quad (5.0 * 10^(6)\; {\rm m\cdot s^(-1)})\\ &\quad\quad (3.0 * 10^(-4)\; {\rm N\cdot s \cdot C^(-1) \cdot m^(-1)}) \\ &\approx 1.0 * 10^(-3)\; {\rm N}\end{aligned}`.