Answer to Question #96808 in Mechanics | Relativity for Sofea

Question #96808
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
1
Expert's answer
2019-10-21T13:35:00-0400

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 = E\vec E


vector of Electric field = B\vec B


Charge in an Electro magnetic field = q


Force = F\vec F


It is given by the Lorentz equation , that is

F=qE+q (ν XB)\vec F = q \vec E + q \space (\vec \nu \space X \vec B)


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



So, the Lorentz equation becomes

F=q (ν XB)\vec F = q \space (\vec \nu \space X \vec B)


Given information,

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


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





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

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


Here,θ=90Here, \theta = 90


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



Force is perpendicular to both V and B\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 =

mν2r\frac {m\nu^2} {r}


Here r is the radius of the circle.



qVB=mν2rq V B = \frac {m\nu^2} {r}

r=mνqBr = \frac {m \nu} {qB}

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

r=mνqBr = \frac {m \nu} {qB}


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