Question

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference of 1.2 x 106 V and then enters a uniform magnetic field whose strength is 2.2 T. The alpha particle moves perpendicular to the field. Calculate (a) the speed of the alpha particle, (b) the magnitude of the magnetic force exerted on it, and (c) the radius of its circular path.

Answer 1

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm