Answer to Question #85352 in Electricity and Magnetism for Vaibhavi Desai

Question

An alpha particle of charge 3.2×10^-19C and mass of 6.7×10^-27kg is fired into a uniform magnetic field of magnetic 0.24T north with a velocity of 8×10^6 ms^-1 to the east. Calculate:

a) The force on the alpha particle
b) The diameter of the subsequent circular motion.

Answer 1

a) We can find the magnetic force on the alpha particle from the formula:

$$F = qvBsin \theta,$$

here, $q$ is the charge of the alpha particle, $v$ is the velocity of the alpha particle, $B$ is the magnetic field and $\theta$ is the angle between the directions of $v$ and $B$.

Then, we get (the angle between the velocity and the direction of the magnetic field is $90^{\circ}$):

$$F = 3.2 \cdot 10^{-19} C \cdot 8 \cdot 10^6 ms^{-1} \cdot 0.24 T \cdot sin90^{\circ} = 6.14 \cdot 10^{-13} N.$$

b) There are two forces that act on the alpha particle when it moves in the uniform magnetic field: the magnetic force and the radial force (this one is required to keep the alpha particle moving in a circle). So, using the Newton’s second law of motion we can write:

$$qvB = \dfrac{mv^2}{r},$$

here, $q$ is the charge of the alpha particle, $v$ is the velocity of the alpha particle, $B$ is the magnetic field, $m$ is the mass of the alpha particle and $r$ is the radius of the alpha particle's circular orbit.

Then, from this formula we can find the radius of the alpha particle's circular orbit and, hence, the diameter of the subsequent circular motion:

$$d = 2r = 2 \dfrac{mv}{qB} = \dfrac{2 \cdot 6.7 \cdot 10^{-27} kg \cdot 8 \cdot 10^6 ms^{-1}}{ 3.2 \cdot 10^{-19} C \cdot 0.24 T} = 1.39 m.$$

Answer:

a) $F = 6.14 \cdot 10^{-13} N.$

b) $d = 1.39 m.$