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Physics Question on Moving charges and magnetism

A cyclotron is operated at an oscillator frequency of 12 MHz12\, MHz and has a dee radius R=50 cmR = 50\,cm . What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron ?

A

0.72 T0.72\,T

B

0.65 T0.65\,T

C

0.39 T0.39\,T

D

0.12 T0.12\,T

Answer

0.72 T0.72\,T

Explanation

Solution

Given, f=12 MHz=12×10−6 Hzf= 12 \,MHz = 12\times 10^{-6}\, Hz
R=50 cm=520×10−2 mR = 50\,cm= 520 \times 10^{-2}\, m
B=2πmfqB = \frac{2\pi mf}{q}
=(12×106)×2×π×(1.67×10−27)(1.6×10−19)= \frac{(12\times 10^{6}) \times 2 \times \pi \times (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})}
=0.788≈0.72 T= 0.788 \approx 0.72\,T