Question

A Particle whose is twice the mass of Proton and Charge is four times the charge of proton is allowed through a uniform magnetic field field of strength 1.7 mT and is pespendicular to the velocity of the particle find the radius and time period of revolution of the particle if the velocity of the particle is 2.5X10^4m/s.​ 1 SEE ANSWER ADD ANSWER

Answer 1

Answer:

An external magnetic field imposes a magnetic force on a charged particle moving through it. The necessary condition for the existence of this force is that the charged particle should not be at rest. This force is always perpendicular to the direction of the magnetic field and the velocity of the particle.

Answer and Explanation:

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Given:

The mass of the particle is

m

=

2

m

p

The charge on the particle is

q

=

4

q

p

The magnetic field is {eq}B = 1.7\times...