1) The magnetic force exerted on the proton, since it is moving perpendicular to the magnetic field, is
[tex]F=qvB[/tex]
where q is the proton charge, v its velocity and B the intensity of the magnetic field.
Since the proton is moving by circular motion, this force is equal to the centripetal force:
[tex]F=qvB=ma_c[/tex]
where m is the proton mass and ac is the centripetal acceleration.
Substituting the data of the problem and re-arranging the formula, we find:
[tex]B= \frac{ma_c}{qv}= \frac{(1.67 \cdot 10^{-27} kg)(1.90 \cdot 10^{13} m/s^2)}{(1.6 \cdot 10^{-19} C)(1.80 \cdot 10^7 m/s)}= 0.011 T[/tex]
so, the magnitude of the magnetic field is 0.011 T.
2) Direction:
the direction of the magnetic field can be found by using the right-hand rule.
Let's take:
- the index finger as the direction of the velocity (positive z-direction)
- the middle finger as the direction of the magnetic field
- the thumb as the direction of the force (which has same direction as the acceleration) (positive x-direction)
we can see that by using the right hand, the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction.
