Answer:
0.0625 T
Explanation:
When a charged particle is moving in a region with a magnetic field, the particle experiences a force given by the equation
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is its velocity
B is the magnitude of the magnetic field
[tex]\theta[/tex] is the angle between the direction of the velocity and of the field
For the proton in this problem, we have:
[tex]F=2.0\cdot 10^{-18}N[/tex] is the force experienced
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]v=200 m/s[/tex] is its velocity
[tex]\theta=90^{\circ}[/tex], since the proton enters the region perpendicular to the magnetic field
Therefore, we can re-arrange the equation to find B:
[tex]B=\frac{F}{qvsin \theta}=\frac{2\cdot 10^{-18}}{(1.6\cdot 10^{-19})(200)(sin 90^{\circ})}=0.0625 T[/tex]
