Question

The oscillator frequency of a cyclotron is 10 MHz what should be the operating magnetic field accelerating proton?

• A. 0.156 T
• B. 0.256 T
• C. 0.356 T
• D. 0.656 T

Answer 1

The correct option is D 0.656 T

It is given that,

Frequency, $\nu =10MHz={10}^{7}Hz$

Mass of proton, $m=1.6\times {10}^{-27}Kg$

Charge $q=1.6\times {10}^{-19}$

Magnetic field,

$B=\frac{2\pi m\nu }{q}$

$B=\frac{2\times 3.14\times 1.67\times {10}^{-27}\times {10}^7}{1.6\times {10}^{-19}}$

$B=\frac{10.4876}{1.6}$

$B=0.656T$