First slide

Force on a charged particle moving in a magnetic field

Question

A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be

Moderate

A

1.5 MeV
B

1 MeV
C

4 MeV
D

0.5 MeV

Solution

The kinetic energy acquired by a charged particle in a uniform magnetic field B is
$K = \frac{q^{2} B^{2} R^{2}}{2 m} (as R = \frac{mv}{qB} = \frac{\sqrt{2 mK}}{qB})$
where q and m are the charge and mass of the particle and R is the radius of circular orbit.
$∴ The kinetic energy acquired by proton is$
$K_{p} = \frac{q_{p}^{2} B^{2} R_{p}^{2}}{2 m_{p}}$
$and that by the alpha particle is$
$K_{α} = \frac{q_{α}^{2} B^{2} R_{α}^{2}}{2 m_{α}}$
$Thus, \frac{K_{α}}{K_{p}} = {(\frac{q_{α}}{q_{p}})}^{2} (\frac{m_{p}}{m_{α}}) {(\frac{R_{α}}{R_{p}})}^{2}$
$or K_{α} = K_{p} {(\frac{q_{α}}{q_{p}})}^{2} (\frac{m_{p}}{m_{α}}) {(\frac{R_{α}}{R_{p}})}^{2}$
$Here, K_{p} = 1 MeV, \frac{q_{α}}{q_{p}} = 2, \frac{m_{p}}{m_{α}} = \frac{1}{4}$ $and \frac{R_{α}}{R_{p}} = 1$
$∴ K_{α} = (1 MeV) (2)^{2} (\frac{1}{4}) (1)^{2} = 1 MeV$

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