Magnetic Effects Of Current Question 413
Question: A cyclotron is operated at an oscillator frequency of 24 MHz and has a dee radius $ R=60cm $ . What is magnitude of the magnetic field B (in Tesla) to accelerate deuterons $ (mass=3.34\times {{10}^{-27}})kg $ ?
Options:
A) 9.5
B) 7.2
C) 5.0
D) 3.2
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given: $ f=24\times 10^{6}Hz,R=0.60m $ We know that, $ R=\frac{mv}{qB} $
$ \therefore ,B=\frac{mv}{qR} $ ?..(i) where $ v=\omega R=2\pi fR $ $ =2\pi \times 24\times 10^{6}\times 0.60=9.04\times 10^{7}m/s $ From (i), $ B=\frac{(3.34\times {{10}^{-27}})(9.04\times 10^{7})}{1.6\times {{10}^{-19}}\times 0.60}=3.2T $ ,
BETA
