Question

An alpha particle, which is a helium nucleus, has a mass of 6.64x10^-27 kg. It is traveling horizontally at 42.3 km/s when it enters a uniform, vertical 1.60 T magnetic field. What is the diameter of the path followed by the particle?

Answer 1

Given:

• Mass, m = 6.64 x 10⁻²⁷ kg

,

• Velocity = 42.3 km/s

,

• Magnetic field = 1.60 T

Let's find the diameter of the path followed by the particle.

To find the path, apply the formula to first find the radius:

`R=(mv)/(QB)`

Where:

m = 6.64 x 10⁻²⁷ kg

v is the velocity in m/s = 42.3 x 1000 = 42300 m/s

Q = 1.6 x 10⁻¹⁹ x 2

B = 1.60 T

Plug in values and solve for the radius, r:

`\begin{gathered} R=(6.64*10^(-27)*42300)/((3.2*10^(-19))*1.60) \\ \\ R=5.48578*10^(-4)\text{ m} \end{gathered}`

Therefore, the diameter will be:

`\text{ diameter = radius x 2 = \lparen5.48578*10}^(-4))*2=0.00101\text{ m}\approx1.01\text{ mm}`

Therefore, the diameter of the path followed by the particle is 1.01 mm.

ANSWER:

1.01 mm