Question

A high energy proton from the sun enters the Earths magnetic field at a speed of 6.0x10^6 m/s perpendicular to the field lines. The field is uniform and with strength B=4.5x10^-5 T.

a) What is the magnetic force on the proton when it first enters the B-field?

b) Calculate the radius of the protons path in the B field.

c) determine the frequency and the period of the protons orbit.

Answer 1

a). Since velcoity of proton and Magnetic Field are perpendicular to each other, then

$F= q(\vec{v} \times \vec{B})= qvBsin90^{\circ}= qvB$

$F= (1.6 \times 10^{-19})(6 \times 10^6)(4.5 \times 10^{-5})$

   $F= 43.2 \times 10^{-18} N$

b), Now since the proton will move in a circular path then above magnetic force will be equal to centripetal force acting on it. Thus

$r= (1.67 \times 10^{-27})(6 \times 10^6)/(1.6 \times 10^{-19})(4.5 \times 10^{-5})$

   $r= 1.391 \times 10^{3} m$

c). Also then Time Period of revolution, $T= 2\pi r/v$

   $T= 2(3.14) (1.391 \times 10^3)/(6 \times 10^6)$

   $T= 1.455 \times 10^{-3} s$

and frequency of revolution is $f=1/T= 1/(1.455 \times 10^{-3})= 687.28 s^{-1}$