The radius of circular path of an electron when subjected to a perpendicular magnetic field is
A . $\dfrac{Be}{mv}$
B . $\dfrac{me}{Bv}$
C . $\dfrac{mB}{ev}$
D . $\dfrac{mv}{Be}$

Hint:In order to answer this question, we have to use the concept that a charged particle moving in a magnetic field experiences a magnetic force that occasionally causes it to deflect. The charge, speed, and magnetic field strength of the particle all affect how strong the magnetic force is.






Formula used:
The magnetic force:
F=qvB; here, q denotes the charge, v the particle's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.




Complete answer:
The centripetal force counterbalances the magnetic pull that acts on a moving particle perpendicular to its velocity. The equilibrium formula states the following:
F(magnetic force)=${F_c}$(centripetal force) - (i)
The centripetal force here operating on the electron is known as ${F_c}$. We are aware that a electron is subject to the force, F, of a magnetic field represented by,
F=evB – (ii)
Here, e denotes the charge of the electron, v the electron's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$ - (iii)
Here m is the mass of the electron, v is the velocity of the electron, and r is the radius.
We will now substitute equation (iii) and (ii) in equation (i) to find the value of r.
$evB=\dfrac{m{{v}^{2}}}{r}$
$r=\dfrac{mv}{eB}$
The correct answer is D.






Note:A force is applied to a particle when the magnetic field and particle are moving perpendicular to one another. If the magnetic field and the particle are moving in the same direction, there is no force exerting on the particle.