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A deuteron of kinetic energy \( 50 \, \text{keV} \) is describing a circular orbit of radius \( 0.5 \, \text{metre} \) in a plane perpendicular to the magnetic field \( \vec{B} \). The kinetic energy of the proton that describes a circular orbit of radius \( 0.5 \, \text{metre} \) in the same plane with the same \( \vec{B} \) is
- 25 keV
- 50 keV
- 200 keV
- 100 keV
Correct answer: 100 keV
Solution
The radius of a charged particle in a magnetic field is given by \( r = \frac{mv}{qB} \). For the same radius and magnetic field, \( \frac{mv}{q} \) must be the same. Since kinetic energy \( KE = \frac{1}{2}mv^2 \), we find that \( KE \propto \frac{1}{m} \). The mass of a deuteron is twice that of a proton, so the proton's kinetic energy will be twice that of the deuteron: \( 2 \times 50 \, \text{keV} = 100 \, \text{keV} \).
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