1.

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.​

Answer»

ANSWER:

r=0.5678m,w=0.8×10^-6

Explanation:

r=mv/qB                               | w=2πm/qB

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PUT VALUE and SOLVE the QUESTION



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