A proton of a mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of a particle (mass 4m and charge +2e) so that it can revolve in the path of same radius?

1 MeV

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4 MeV

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2 MeV

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0.5 MeV

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Solution

The correct option is A 1 MeV
$r = \frac{\sqrt{2 m k}}{q B}$

$K = \frac{k q^{2}}{m}$ where k is constant

$\frac{K_{p r o t o n}}{K_{p a r t i c l e}} = \frac{(q_{P})^{2} 4 m_{p}}{(q_{a})^{2} m_{p}} = \frac{(q_{P})^{2} m_{a}}{(2 q_{p})^{2} m_{p}} = 1 M e V$

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A proton of a mass m and charge +e is moving in a circular orbit of a magnetic field with energy 1 MeV. What should be the energy of a particle (mass 4m and charge +2e) so that it can revolve in the path of same radius?

The correct option is A 1 MeVr=√2mkqBK=kq2m where k is constantKprotonKparticle=(qP)24mp(qa)2mp=(qP)2ma(2qp)2mp=1MeV

The correct option is A 1 MeV
$r = \frac{\sqrt{2 m k}}{q B}$

$K = \frac{k q^{2}}{m}$ where k is constant

$\frac{K_{p r o t o n}}{K_{p a r t i c l e}} = \frac{(q_{P})^{2} 4 m_{p}}{(q_{a})^{2} m_{p}} = \frac{(q_{P})^{2} m_{a}}{(2 q_{p})^{2} m_{p}} = 1 M e V$