Question

At a particular instant, a proton is traveling west to east with a kinetic energy of 12 keV. Earth's magnetic field has a horizontal component of 4.4 ✕ 10⁻⁵ T north and a vertical component of 2.0 ✕ 10⁻⁵ T down.

What is the radius of curvature of the path (in m)?

Answer 1

m = mass of proton = 1.67 x 10^-27 kg

v = speed of proton = ?

KE = kinetic energy = 12 keV = 12 x 1000 x 1.6 x 10^-19 = 1.92 x 10^-15 J

kinetic energy of proton is given as

KE = (0.5) m v²

1.92 x 10^-15 = (0.5) (1.67 x 10^-27) v²

v = 1.52 x 10⁶ m/s

q = 1.6 x 10^-19 C

B_y = vertical component of magnetic field = 2 x 10^-5 T

B_x = horizontal component of magnetic field = 4.4 x 10^-5 T

net magnetic field is given as

B = sqrt(B_x² + B_y²) = sqrt((4.4 x 10^-5)² + (2 x 10^-5)²) = 4.83 x 10^-5 T

radius of curvature is given as

r = mv/(qB)

r = (1.67 x 10^-27) (1.52 x 10⁶)/((1.6 x 10^-19) (4.83 x 10^-5))

r = 328.5 m