Soaring Crane
- 461
- 0
1) For a particle of mass m and a charge of q moving in a circular path in a magnetic field B, show that its kinetic energy is proportional to r^2, the square of the radius of the curvature of the path.
I started off with the definition of KE: (mv)^2/2 and the fact that a = v^2/r, where F = m(v^2/r).
Then, I thought;
mv^2 = 2*KE and mv^2 = F*r so r = (2KE)/F so r^2 = (4KE^2)/ F^2, but this is probably wrong. What should I do if so?
Thanks.
I started off with the definition of KE: (mv)^2/2 and the fact that a = v^2/r, where F = m(v^2/r).
Then, I thought;
mv^2 = 2*KE and mv^2 = F*r so r = (2KE)/F so r^2 = (4KE^2)/ F^2, but this is probably wrong. What should I do if so?
Thanks.