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  4. A deuteron of kinetic energy 50 KeV describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be

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Q. A deuteron of kinetic energy $50\,KeV$ describes a circular orbit in a magnetic field .The kinetic energy of a proton that describes a circular orbit of the same radius in the same field would be

Solution:

$K E=\frac{B^{2} \,q^{2} \,r^{2}}{2 m}$
$K E_{2}=K E_{1} \times \frac{m_{1}}{m_{2}}$
$=50 \times \frac{2 m}{m}=100 \,keV$