Question

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV. The beam is then entered into a region between two parallel metal plates with potential difference 120 V and a separation 8 mm, perpendicular to the direction of the field. What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates?

Answer 1

The magnetic field required required for the beam not to be deflected is
`B = 0.0036T`

Step-by-step explanation:

From the question we are told that

The charge on the particle is
`q = +2e`

The mass of the particle is
`m = 6.64 *10^(-27) kg`

The potential difference is
`V_a = 1.8 kV = 1.8 *10^(3) V`

The potential difference between the two parallel plate is
`V_b = 120 V`

The separation between the plate is
`d = 8 mm = (8)/(1000) = 8*10^(-3)m`

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

`KE_b = PE_b`

Generelly

`KE_b = (1)/(2) m v^2`

And
`PE_b = q V_a`

Equating this two formulas

`(1)/(2) mv^2 = q V_a`

making v the subject

`v = \sqrt{(q V_a)/(2 m) }`

Substituting value

`v = \sqrt{( 2* 1.602 *10^(-19) * 1.8 *10^(3))/(2 * 6.64 *10^(-27)) }`

`v = 41.65*10^4 m/s`

Generally the electric field between the plates is mathematically represented as

`E = (V_b)/(d)`

Substituting value

`E = (120)/(8*10^(-3))`

`E = 15 *10^3 NC^(-1)`

the magnetic field is mathematically evaluate

`B = (E)/(v)`

`B = (15 *10^(3))/(41.65 *10^4)`

`B = 0.0036T`