Question

An alpha particle consists of two protons and two neutrons, each with a mass of about 1.67 x 10-27 kilograms. Consider an alpha particle traveling at 9.26 x 106 meters per second perpendicular to a magnetic field of strength 2.84 teslasa. What magnetic force acts on the alpha particle?b. Since the magnetic force on the alpha particle is always perpendicular to its velocity, the magnetic force acts as a centripetal force. What is the radius of curvature of the path of the alpha particle?

Answer 1

Given:

• Mass of each, m = 1.67 x 10⁻²⁷ kg

,

• Velocity, v = 9.26 x 10⁶ m/s

,

• B = 2.84 teslas

Let's sole for the following:

• (a). What magnetic force acts on the alpha particle?

Apply the formula:

`F=qvB`

Where:

Charge of proton = 1.609 x 10⁻¹⁹ C

Charge of neutron = 0

Thus, we have:

`\begin{gathered} F=(2*1.609*10^(-19))*(9.26*10^6)*2.84 \\ \\ F=8.46*10^(-12)\text{ N} \end{gathered}`

The magnetic force that acts on the alpha particle is 8.46 x 10⁻¹² N.

• (b). Let's find the radius of curvature of the path of the alpha particle?

To find the radius, apply the formula:

`F=(mv^2)/(r)`

Where:

F is the force

m is the total mass

v is the velocity

r is the radius

Rewrite the formula for r, plug in the values and solve.

We have:

`\begin{gathered} r=(mv^2)/(F) \\ \\ r=((4*1.67*10^(-27))*(9.26*10^6)^2)/(8.46*10^(-12)) \\ \\ r=(5.7279*10^(-13))/(8.46*10^(-12)) \\ \\ r=0.0677 \\ \\ r=6.77*10^(-2)\text{ m} \end{gathered}`

The radius of the curvature of the path is 6.77 x 10⁻² m.

ANSWER:

(a). 8.46 x 10⁻¹² N.

(b). 6.77 x 10⁻² m.