Electric Charge in a Magnetic Field

AI Thread Summary
In a magnetic field, a charged particle of mass m and charge q moves in a circular path, where its kinetic energy (KE) is derived from the relationship between force and acceleration. The equations F = mv²/r and F = Bqv are crucial for understanding the forces acting on the particle. The discussion explores how to express the radius r in terms of kinetic energy, leading to the conclusion that r² is proportional to KE when considering the forces involved. The relationship between the magnetic field strength B and distance R is also examined, noting that B typically varies with r. Ultimately, the goal is to establish that the kinetic energy of the particle is proportional to the square of the radius of its circular path.
Soaring Crane
Messages
461
Reaction score
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.
 
Physics news on Phys.org
What do you know about the strength of magnetic fields in relation to distance R?
 
If B is constant, then r is constant??
 
Well typically one would say that backwards, since B is a function of r. What is B as a function of R?

What does B have to be equal to to put the particle in uniform circular motion? Think about the force involved in circular motion, and the force of a magnetic field.
 
Soaring Crane said:
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.
Remember:
F=mv²/r
and F=Bqv

also v=(2pi.r)/t
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top