A) 0.016 T
B) 0.16 T
C) 16 T
D) 1.6T
Correct Answer: D
Solution :
A cyclotron accelerates charged particles with a high-frequency, alternating voltage (potential difference). A perpendicular magnetic field causes the particles to go almost in a circle. Also a high frequency alternating voltage applied across the 'D1 electrodes alternately attracts and repels charged particles. The centripetal force is provided by the transverse magnetic field B, and the force on a particle travelling in a magnetic field (which causes it to curve) is equal to BqV. So, \[\frac{m{{v}^{2}}}{r}=Bqv\]where, m is mass, q is charge, v is velocity and r is radius. Also \[v=r\omega \] \[\therefore \] \[\omega =\frac{Bq}{m}\] Since, \[\omega =2\pi n,\]we have \[2\pi n=\frac{Bq}{m}\] \[\Rightarrow \] \[B=\frac{2\pi n\times m}{q}\] \[=\frac{2\times \pi \times 12\times {{10}^{6}}\times 3.3\times {{10}^{-27}}}{1.6\times {{10}^{-19}}}\] \[\approx 1.6\,\text{T}\]
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