Question

A proton is performing circular motion with radius 0.5 m under effect of normal magnetic field 1 Telsa. Kinetic energy of proton is:-

Answer 1

Hint: Kinetic energy is the sort of energy that an item or particle has as a result of its motion. When activity, which transfers energy, is performed on an item by exerting a net force, the object accelerates and obtains kinetic energy. Kinetic energy is a characteristic of a mobile object or particle that is affected by both its speed and its mass.
It is represented by the formula that is K.E=$\dfrac{1}{2}m{{v}^{2}}$$\cdots (1)$

Complete answer:
In this given question proton is moving in a circular motion with radius 0.5m under the effect of magnetic field. We need to find out the kinetic energy of the proton

We know that,
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Where $m$is the mass and $v$is the velocity of the particle. Here in this case particle is proton.

As we know that, radius of the charge in the magnetic field$R=\dfrac{mv}{qB}$, where q is the charge and $B$ is the magnetic field.
$\Rightarrow $$mv=(qB)R$

Now, putting the value of $mv$ in equation (1)
$K.E=\dfrac{1}{2}\dfrac{{{(mv)}^{2}}}{m}$
$\Rightarrow \dfrac{1}{2}\dfrac{{{(qBR)}^{2}}}{m}$

Given value of, $B=1tesla$, $q=1.6\times {{10}^{-19}}$, $R=0.5$ and mass of the proton is$1.6\times {{10}^{-27}}$

Now put this values in the above formula,
$K.E=\dfrac{1}{2}\dfrac{{{(1.6\times {{10}^{-19}}\times 1\times 0.5)}^{2}}}{1.6\times {{10}^{-27}}}$
$\Rightarrow \dfrac{25}{2}\times {{10}^{6}}$=$12.5\times {{10}^{6}}eV$
$K.E=12.5meV$

Therefore, kinetic energy of the proton in the circular path is $12.5meV$.

Note: In order to answer these questions, we must first determine the mass and charge of these particles. Also, grasp the relationship between the radius of a circular pathway and the magnetic field. Concept and formula of the kinetic energy must be known.