Question

A cyclotron’s oscillator frequency is $n$ and radius of the dees is $r$. The operating magnetic field $(B)$ for accelerating protons of charge $q$ and kinetic energy of protons produced by the accelerator is respectively ($m$ and $v$ be the mass and velocity of proton)

• A. $\dfrac{2\pi nm}{q},\; \dfrac{qvBr}{2}$
• B. $\dfrac{\pi nm}{q},\; \dfrac{qvBr}{2}$
• C. $\dfrac{2\pi nm}{q},\; qvBr$
• D. $\dfrac{4\pi nm}{q},\; \dfrac{qvBr}{2}$

Hint:

Cyclotron frequency is independent of velocity and radius of particle.

Answer 1

The Correct Option is A

Step 1: Cyclotron frequency relation.
The cyclotron frequency is given by:
\[ n = \frac{qB}{2\pi m} \]
Step 2: Expression for magnetic field.
\[ B = \frac{2\pi nm}{q} \]
Step 3: Kinetic energy of accelerated proton.
Radius of circular path:
\[ r = \frac{mv}{qB} \] \[ v = \frac{qBr}{m} \]
Step 4: Kinetic energy expression.
\[ K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{qBr}{m}\right)v = \frac{qvBr}{2} \]
Step 5: Conclusion.
The correct pair is $\left(\dfrac{2\pi nm}{q},\; \dfrac{qvBr}{2}\right)$.