CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Electric Field

A charge part...

Question

A charge particle of mass $m$ and charge $q$ entered int magnetic field $B$ normally after accelerating by potential difference $V$ . Calculate radius of its circular path.

Open in App

Solution

Kinetic energy of the particle $K . E = q V$
So, $m v = \sqrt{2 m K . E} = \sqrt{2 m q V}$
Radius of circular path $R = \frac{m v}{q B}$
$⟹ R = \frac{\sqrt{2 m q V}}{q B}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. A charged particle of mass

m

q

is accelerated through a potential difference of

V

. It enters a region of uniform magnetic field

B

, which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by -

Q. A charged particle of mass m and charge

q

is accelerated through a potential difference of

V

volts.It enters a region of uniform magnetic field

B

which is directed perpendicular to the direction of motion of the particle.The particle will move on a circular path of radius

Q. A charged particle revolves in circular path in uniform magnetic field after accelerating by a potential difference of V volts.Choose the correct options if V is doubled.

Q. An electron of mass

m

is accelerated through a potential difference of

V

and then it enters a magnetic field of induction

B

normal to the lines. Then, the radius of the circular path is

Q. A particle of charge q, mass m and velocity v moves in a circular path inside the dees of a cyclotron. If the magnetic field exerted on the particle is B, calculate the radius R of the particles path.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

The Electric Field

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon