Question

What is the radius of the circular path? ANY HELP(: A proton moves at a speed of 8.4x10^4 m/s as it passes through a magnetic field of 12 mT.

Answer 1

Well start off with the force on a moving charge through a magnetic field.

F=q(vxB)
(V is crossed With B)

Using the definition of a cross product this reduces to F = QvBSin(theta)

Assuming the magnetic filed is perpendicular with the velocity the sine of angle between them is 1.

Equating this with the centripetal force for circular motion yields

mv2/r=qvB


Solving for r you get

r=(mv)/(qb)


From there just plug in the numbers.

Hope this Helps :)

Answer 2

There is a typo where i equated the forces it should be m(v^2)/r :)

Answer 3

What should go where when plugging in? >.<

Answer 4

Where;

m= mass of the proton
B= Magnetic field
v= velocity of the proton