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Question: An electron having mass \(m\) and kinetic energy \(E\) enter in the uniform magnetic field \(B\) , p...

An electron having mass mm and kinetic energy EE enter in the uniform magnetic field BB , perpendicularly, then its frequency will be:
a) eEqVB\dfrac{{eE}}{{qVB}}
b) 2πmeB\dfrac{{2\pi m}}{{eB}}
c) eB2πm\dfrac{{eB}}{{2\pi m}}
d) 2meBE\dfrac{{2m}}{{eBE}}

Explanation

Solution

Uniform magnetic field is the magnetic force experienced by an object when magnetic lines are parallel. Frequency is something that occurs in a unit time.it is given by the formula: F=1TF = \dfrac{1}{T} , where F=F = frequency, T=T = time.

Complete step by step answer:
Uniform magnetic field: it is defined as the magnetic force that is experienced by an object when the magnetic lines are parallel and all are at all the same points.
For example: the strength of the bar magnet is greater towards the end of poles.
Velocity is defined as the speed of a thing at a particular direction.
The magnetic force is given by the formula: F=qVBF = qVB
Where, F=F = magnetic force
q=q = charge
V=V = particle velocity
B=B = magnetic field
The centripetal force is given by the formula: qVB=mv2rqVB = \dfrac{{m{v^2}}}{r}
Where, m=m = mass
v=v = velocity
r=r = radius.
So, if the velocity of the particle is perpendicular to the magnetic field, then the particle moves in a circular motion.
Then the centripetal force that is experienced by an electron will be given as qVBqVB
Applying this, we will get qVB=mv2rqVB = \dfrac{{m{v^2}}}{r}…..1
qB=mvrqB = \dfrac{{m{v^{}}}}{r} ….2
Now as we know, v=ωRv = \omega R
Substituting this value, in the above equation 2 we get,
qB=mωqB = m\omega
Therefore, ω=qBm\omega = \dfrac{{qB}}{m}…3
We know that, ω=2πT\omega = \dfrac{{2\pi }}{T} , Substitute this value in equation 3 we get,
qBM=2πT\dfrac{{qB}}{M} = \dfrac{{2\pi }}{T} ….4
After all this we will see about the frequency.
It is defined as the number of occurrences per unit time. It is given by the formula F=1TF = \dfrac{1}{T}
But the value of 1T=qB2πm\dfrac{1}{T} = \dfrac{{qB}}{{2\pi m}}
Substituting this value in equation 4 we get,
1T=eB2πm\dfrac{1}{T} = \dfrac{{eB}}{{2\pi m}}
Therefore, F=eB2πmF = \dfrac{{eB}}{{2\pi m}}
Where, F=F = frequency
e=e = energy
B=B = magnetic field
m=m = mass of the particle

So, the correct answer is option C) eB2πm\dfrac{{eB}}{{2\pi m}}

Note: Magnetic force is perpendicular to the direction it travels. Therefore, the charged particles follow a curved path in the magnetic field. The magnetic field does not depend on the velocity and the radius of the circular path.