A magnetic field will deflect a charged particle perpendicular to its direction of motion. An electron experiences a radial acceleration of 8.4E14 m/s^2 in one such field.

Question

What is its speed (in meters/second) if the radius of its curved path is 0.69 m?

Answer 1

When a particle is subjected to a uniform external magnetic field (which provides an external force), the velocity and direction of the magnetic field are perpendicular. (This is significant because it means the magnetic field does no work on the particle so the speed and kinetic energy of the particle are constant. The velocity is not constant since the direction of the particle is changing but the speed itself is constant. That is an important distinction.)

As such, the magnetic field provides ONLY a radial, or centripetal, force on the particle, which can be written as:

F= mv²/r.

From Newton’s 2nd law, we know that F = ma so this can be rewritten as:

>ma = mv²/r

>a = v²/r

Rearrange:

>v² = ar

>v = sqrt(ar)

>v = sqrt((8.4 * 10^14)*(0.69))

>v = 2.4 * 10^7 m/s