A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040 T at a right angle to the field. What is the radius of the proton's resulting orbit

r = 0.114 m

Explanation:

To find the speed of the proton, from conservation of energy, we know that

KE = PE

Thus, we have;

(1/2)mv² = qV

Where;

V is potential difference = 1kv = 1000V

q is charge on proton which has a value of 1.6 x 10^(-19) C

m is mass of proton with a constant value of 1.67 x 10^(-27) kg

Let's make the velocity v the subject;

v =√(2qV/m)

v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))

v = 4.377 x 10^(5) m/s

Now to calculate the radius of the circular motion of charge we know that;

F = mv²/r = qvB

Thus, mv²/r = qvB

Divide both sides by v;

mv/r = qB

Thus, r = mv/qB

Value of B from question is 0.04T

Thus,

r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)

r = 0.114 m

r = 8.76 m