A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (mproton=1.67 × 10-27 kg, e = 1.60 × 10-19 C)

Explanation:

In the region of potential difference , protons will acquire velocity which can be found as follows

1/2 m v² = Ve , V is potential diff , v is velocity attained , m is mass of proton , e is charge on proton

v² = 2Ve / m

For circular path of proton in a  magnetic field

mv² / r = Bev    r is radius of circular path.

r = mv / Be

r² = m²v² / B²e²

= 2Ve/m x m²/ B²e²

= 2Vm / B²e

= 2 x 1000 x 1.67 x 10⁻²⁷ /( .04² x 1.6 x 10⁻¹⁹ )

= .013

r = .114 m

= 11.4 cm

thermal energy is the answer