# Motion of Electron in Magnetic Field

When a charge particle ‘q’ moves in magnetic field ‘B’ with velocity ‘v’ then the force experienced by the charge is

$${\vec{F} = q(\vec{v}\times \vec{B})}$$

F = Bqvsinθ

Where ‘θ’ is the angle between and $\vec{v}$ and $\vec{B}$. The direction of force is perpendicular to both v and B.

Derivation:

Let us consider a beam of electrons entering a uniform magnetic field ‘B’ with velocity ‘v’. For simplicity let the direction of the magnetic field ‘B’ is perpendicularly inward the paper and electrons enter the magnetic field as shown in figure.

Motion of electron in Magnetic Field

Since the force is perpendicular to the velocity, the charge describes a circular path and the required amount of centripetal force for circular motion is provided by magnetic force Bevsinθ. Since v is perpendicular to B so sinθ = 90º = 1.

$${\frac{mv^{2}}{r} = BeV}$$
or, $${R =\frac{mv}{Be}}$$
$${When \theta \neq 90^{\circ}}$$

$${r = \frac{mvsin\theta }{Be}}$$

This is the required equation for the motion of electron in magnetic field.

Do you like this article ? If yes then like otherwise dislike :