Explanation:
It is given that,
Speed of proton, [tex]v=5\times 10^7\ m/s[/tex]
Magnetic force, [tex]F=1.7\times 10^{-16}\ N[/tex]
(1) The strength of the magnetic field if there is a 45º angle between it and the proton’s velocity. The magnetic force is given by :
[tex]F=qvB\ sin\theta[/tex]
[tex]B=\dfrac{F}{qv\ sin\theta}[/tex]
[tex]B=\dfrac{1.7\times 10^{-16}\ N}{1.6\times 10^{-19}\ C\times 5\times 10^7\ m/s\ sin(45)}[/tex]
B = 0.00003 T
or
B = 0.03 mT
The magnitude of Earth's magnitude of 25 to 65 Tesla. The value obtained in part (a) is not consistent with the known strength of the Earth’s magnetic field on its surface.
