A proton moving perpendicular to a magnetic field possesses kinetic energy E. The magnetic field is increased 8 times. But the proton is constrained to move in the path of the same radius. What will be the effect on its energy?

Open in App

Solution

The radius of the path is given by
$r = \frac{m v}{q B} = \frac{\sqrt{2 m E}}{q B}$

When B increases, for r to remain constant, E will increase.

Suggest Corrections

Similar questions

Q. A proton with kinetic energy

8 e V

is moving in a uniform magnetic field. The kinetic energy of a deuteron moving in the same path in the same magnetic field will be:

Q. A proton carrying

1 M e V

kinetic energy is moving in a circular path of radius

R

in uniform magnetic field. What should be the energy of an

α

-particle to describe a circle of same radius in the same field?

A proton with 1 MeV kinetic energy is moving in a circular path of radius R in a uniform magnetic field. What should be the energy of an α - particle to describe a circle of same radius in the same magnetic field?

Q. A proton of mass

m

and charge

q

is moving in a plane with kinetic energy

E

. If there exists a uniform magnetic field

B

, perpendicular to the plane of the motion, the proton will move in a circular path of radius:

Q. A proton of energy

8 eV

is moving in a circular path in a uniform magnetic field. The energy of an

α

- particle moving in the same magnetic field and along the same path will be

Join BYJU'S Learning Program

A proton moving perpendicular to a magnetic field possesses kinetic energy E. The magnetic field is increased 8 times. But the proton is constrained to move in the path of the same radius. What will be the effect on its energy?

The radius of the path is given by r=mvqB=√2mEqB When B increases, for r to remain constant, E will increase.

The radius of the path is given by
$r = \frac{m v}{q B} = \frac{\sqrt{2 m E}}{q B}$

When B increases, for r to remain constant, E will increase.