Question

A proton travels little distance in an electric field, then it enters a crossed magnetic field of 1 T and radius 0.2 m. Find the velocity of the proton.
A. $0.2 \times 10^8 ms^{-1}$
B. $0.2 \times 10^7 ms^{-1}$
C. $0.2 \times 10^6 ms^{-1}$
D. $2 \times 10^8 ms^{-1}$

Answer 1

For a proton, entering a magnetic field, the magnitude of centripetal force of circular motion that it exhibits in the magnetic field is equal to the magnitude of Lorentz Force acting on the proton.

Formula used:
The velocity of a particle of mass m and charge q, in a magnetic field B is
$v = \dfrac{rqB}{m}$,
where r is the radius of circular motion.

Complete step-by-step answer:
Equating the magnitudes of centripetal force (of circular motion) and Lorentz force acting on a proton, we have:
$\dfrac{mv^2}{r} = q v B$
$\implies v = \dfrac{rqB}{m}$.
Now, we are given r = 0.2 m and B = 1 T. And we know for a proton,
$q = 1.6 \times 10^{-19}$ C and
$m = 1.66 \times 10^{-27}$ kg.
Substituting the values in the formula, we get:
$v = \dfrac{ 0.2 m \times 1.6 \times 10^{-19} C \times 1 T}{1.66 \times 10^{-27} kg}$
$\implies v = 0.2 \times 10^8$ m/s.

So, the correct answer is “Option A”.

Additional Information: If the proton or any charged particle enters with a velocity in a magnetic field region, it starts performing a circular motion under the influence of Lorentz force. The direction of velocity and a magnetic field determine the Lorentz Force acting on the particle. If an electric field is also acting such that its direction is perpendicular to the direction of particle velocity, then the particle performs circular motion due to magnetic field and is accelerated due to electric field. Therefore, the trajectory of the particle becomes a helix with changing pitch.
In any situation if the particle entering an electric field has a component of velocity intact in direction parallel to the magnetic field, it will perform helical motion in the magnetic field.

Note: Electric field has been described in the question to acknowledge that proton enters the magnetic field with some initial velocity. It makes a certain angle with the direction of the magnetic field which in our case is just the right angle as the magnetic field is perpendicular to the velocity direction.