Question

A deuteron of kinetic energy $50 \,keV$ is describing a circular orbit of radius $0.5$ metre in a plane perpendicular to the magnetic field $B$. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ metre in the same plane with the same $B$ is

• A. 25 ke V
• B. 50 ke V
• C. 200 ke V
• D. 100 ke V

Answer 1

Answer: (D)

100 ke V

Answer 2

For a charged particle orbiting in a circular path in a magnetic field $\frac{ mv ^{2}}{ r }= Bqv \Rightarrow v =\frac{ Bqr }{ m }$ $or , mv ^{2}= Bqvr$ Also, $E _{ K }=\frac{1}{2} mv ^{2}=\frac{1}{2} Bqvr$ $= Bq \frac{ r }{2} \cdot \frac{ Bqr }{ m }=\frac{ B ^{2} q ^{2} r ^{2}}{2 m }$ For deuteron, $E_{1}=\frac{B^{2} q^{2} r^{2}}{2 \times 2 m}$ For proton, $E_{2}=\frac{B^{2} q^{2} r^{2}}{2 m}$ $\frac{E_{1}}{E_{2}}=\frac{1}{2}$ $\Rightarrow \frac{50 keV }{E_{2}}=\frac{1}{2}$ $\Rightarrow E_{2}=100\, keV .$