Answer to Question #235386 in Electricity and Magnetism for Angie

Question

Question 1 [4 marks]

A proton is accelerated through a potential difference of 3 kV, and then enters a magnetic field of flux density 0.6 T, in a direction perpendicular to the direction of the magnetic field. It travels at right angles to the magnetic field at all times. If the charge-to-mass ratio for the proton is 9.53x107 C kg-1, calculate the radius of the proton’s circular orbit.

Answer 1

$V= 3 \;kV = 3000 \;V \\ B = 0.6 \;T \\ \frac{g}{m} = 9.53 \times 10^7 \;c/kg$

Radius of proton’s circular orbit

$r=\frac{1}{B} \times \sqrt{\frac{2mV}{q}} \\ = \frac{1}{B} \times \sqrt{\frac{2V}{q/m}} \\ r = \frac{1}{0.6} \times \sqrt{\frac{2 \times 3000}{9.53 \times 10^7}} \\ = 1.3224 \times 10^{-2} \;m \\ = 1.32 \;cm$