The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
![F=qvBsin\theta=qvB](https://tex.z-dn.net/?f=F%3DqvBsin%5Ctheta%3DqvB)
since
![\theta = 90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2090%5E%7B%5Ccirc%7D)
, because the velocity v and the force F in this problem are perpendicular, and so also the angle
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
between the velocity and the magnetic field B should be
![90^{\circ}](https://tex.z-dn.net/?f=90%5E%7B%5Ccirc%7D)
.
Let's find the magnitude of the magnetic field; this is given by
![B= \frac{F}{qv}= \frac{5.4\cdot 10^{-14}N}{1.6\cdot 10^{-19}C \cdot 3.7\cdot 10^6 m/s}=0.091 T](https://tex.z-dn.net/?f=B%3D%20%5Cfrac%7BF%7D%7Bqv%7D%3D%20%5Cfrac%7B5.4%5Ccdot%2010%5E%7B-14%7DN%7D%7B1.6%5Ccdot%2010%5E%7B-19%7DC%20%5Ccdot%203.7%5Ccdot%2010%5E6%20m%2Fs%7D%3D0.091%20T%20%20)
To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force
Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T