Question

A cosmic ray proton moving toward the Earth at 5.00×107 m/s experiences a magnetic force of 1.70×10−16 N . What is the strength of the magnetic field if there is a 45º angle between it and the proton’s velocity? (b) Is the value obtained in part (a) consistent with the known strength of the Earth’s magnetic field on its surface?

Answer 1

Step-by-step explanation:

It is given that,

Speed of proton,
`v=5* 10^7\ m/s`

Magnetic force,
`F=1.7* 10^(-16)\ N`

(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 :

`F=qvB\ sin\theta`

`B=(F)/(qv\ sin\theta)`

`B=(1.7* 10^(-16)\ N)/(1.6* 10^(-19)\ C* 5* 10^7\ m/s\ sin(45))`

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.