Question

A particle of mass m and charge q enters a magnetic field B perpendicularly with a velocity v, The radius of the circular path described by it will be
A . ${\dfrac{Bq}{mv}}$
B . ${\dfrac{mq}{Bv}}$
C . ${\dfrac{mB}{qv}}$
D . ${\dfrac{mv}{Bq}}$

Answer 1

Hint:In this question we have to use the concept that when a charged particle moves in a magnetic field, a magnetic force operates on it and occasionally causes it to deflect. The strength of the magnetic force depends on the particle's charge, speed, and the strength of the magnetic field.




Formula used:
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 magnetic force acting on a moving particle perpendicular to its motion is balanced by the centripetal force. The equilibrium formula can be expressed as,
$F={F_c}$ - (i)
In this case, Fc is the centripetal force acting on a charged particle. We are aware that the force, F, acting on a charged particle in a magnetic field is represented by,
F=qvB – (ii)
Here, q denotes the charge, v the particle's speed, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}={\dfrac{m{{v}^{2}}}{r}}$ - (iii)
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.
We will now substitute equation (iii) and (ii) in equation (i) to find the value of r.
$qvB={\dfrac{m{{v}^{2}}}{r}}$
$r={\dfrac{mv}{qB}}$
The correct answer is D.





Note:When a particle and magnetic field are moving relative to one another, a force is exerted on that particle. The force acting on the particle is 0 if the magnetic field and the particle are travelling parallel to each other.