Answer to Question #96808 in Mechanics | Relativity for Sofea

Question

Consider the motion of particle mass m and charge q in an electromagnetic field with
electric field vector is E and the magnetic field vector is B. The force acting on the particle is
given by the Lorentz equation F = qE + qv x B (assuming non-relativistic case, v<<c).

(a) If there is no electric field and the particle enters the magnetic field in a direction
perpendicular to the lines of magnetic flux, show that the trajectory is a circle with radius

Answer 1

Show that the trajectory is a circle with radius

We need to show the trajectory is a circle and find the radius of the circle

Solution:

From the given data = $\vec E$

vector of Electric field = $\vec B$

Charge in an Electro magnetic field = q

Force = $\vec F$

It is given by the Lorentz equation , that is

$$\vec F = q \vec E + q \space (\vec \nu \space X \vec B)$$

(a). If no Electric field is there, means $\vec E = 0$

So, the Lorentz equation becomes

$$\vec F = q \space (\vec \nu \space X \vec B)$$

Given information,

the particle enters the magnetic field ($\vec B$ ), in perpendicular direction to the Lines of Magnetic Flux ($\vec V$ )

It means, $\vec B \space and \space \vec V$ are perpendicular

$$\vec F =q\space (\vec V X \vec B)$$

$$\vec F = q \space V \space B \space sin \space \theta$$

$$Here, \theta = 90$$

$$\vec F = q \space V \space B \space$$

Force is perpendicular to both $\vec V \space and \space \vec B$ and it is in the direction of centre.

Here the magnitude of force is same but directions are different.

These are towards the centre.

So this makes the Trajectory as a Circle.

a force that acts on a body moving in a circular path and is directed towards the centre around which the body is moving

is called a "centripetal Force".

Centripetal force =

$$\frac {m\nu^2} {r}$$

Here r is the radius of the circle.

$$q V B = \frac {m\nu^2} {r}$$

$$r = \frac {m \nu} {qB}$$

Answer: shown that the trajectory is a circle and radius of the circle =

$$r = \frac {m \nu} {qB}$$