Question

A cyclotron’s oscillator frequency is 10 MHz. the operating magnetic field for the cyclotron to accelerate the proton is ( mass of proton $=1.67\times {10}^{-27}kg$

• A. 0.36 T
• B. 0.46 T
• C. 0.56 T
• D. 0.66 T

Answer 1

The correct option is D

0.66 T

$f=\frac{Bq}{2\pi m}\therefore B=\frac{2\pi mf}{q}$

substituting, B = 0.66T