Question

Extra Credit: Show all your work to obtain credit. 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

Answer:

B = 0.036 T

Explanation:

Given:

$m = 6.64*10^-^2^7$

p.d, Va= 1.8 KV = 1800V

Distance btw plates, d= 8mm = 0.008m

$q = 2 * 1.6*10^-^1^9$

Let's use the equation:

$q*Va = \frac{1}{2} mv^2$

Substitute figures in the equation, we have:

$2*1.6*10^-^1^9 * 1800 = \frac{1}{2} * 6.64*10^-^2^7 * v'^2$

Solving for v' we have:

$v' = 41.65 * 10^4$

For electric field between plates, we use the formula :

$E = \frac{V}{d}$

Where V = 120

$E = \frac{120}{0.008}$

$15*10^3 N/C$

The magnitude of magnetic field, B, needed so that the alpha particles emerge undeflected will be given as:

$B = \frac{E}{v'}$

$B = \frac{15*10^3}{41.65*10^4}$

B = 0.036 T