Question

A proton is in moving in a circular orbit radius 14cm I a uniform 0.35T magnetic field perpendicular to the velocity and proton find speed of proton

Answer 1

There are two forces that act on the proton when it moves in the uniform magnetic field: the magnetic force and the radial force (this one is required to keep the proton moving in a circle). So, using the Newton’s Second Law of Motion we can write:

$$qvB=\dfrac{mv^2}{r},$$ $$v=\dfrac{qBr}{m},$$ $$v=\dfrac{1.6\cdot10^{-19}\ C\cdot 0.35\ T\cdot 0.14\ m}{1.67\cdot10^{-27}\ kg}=4.7\cdot10^6\ \dfrac{m}{s}.$$

Answer:

$v=4.7\cdot10^6\ \dfrac{m}{s}.$