Deuterons are accelerated in a cyclotron that has an oscillatory frequency of 107 Hz and a dee radius of 50 cm. The strength of magnetic field needed to accelerate the deuterons is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.31 T
A cyclotron works only when the frequency of the oscillator equals the cyclotron frequency of the particle, the frequency of its circular motion in the magnetic field. This frequency does not depend on the speed or the radius of the path.
\( \displaystyle f = \frac{qB}{2\pi m} \)
\( \displaystyle B = \frac{2\pi m f}{q} \)
A deuteron has the charge of a proton but about twice its mass:
\( q = 1.6\times10^{-19}\,\mathrm{C} \)
\( m = 2 \times 1.67\times10^{-27}\,\mathrm{kg} = 3.34\times10^{-27}\,\mathrm{kg} \)
\( \displaystyle B = \frac{2\pi \times 3.34\times10^{-27}\,\mathrm{kg} \times 10^{7}\,\mathrm{Hz}}{1.6\times10^{-19}\,\mathrm{C}} \)
\( \displaystyle B = \frac{2.10\times10^{-19}}{1.6\times10^{-19}}\,\mathrm{T} \)
\( \displaystyle B \approx 1.31\,\mathrm{T} \)
The dee radius is not needed here; it only decides the maximum energy the deuterons can gain.