Recent content by TSny

If possible, please type your equations using Latex rather than post images of your hand-written work. I'm having difficulty deciphering parts of your diagram and some of the terms in your equations. Using energy is a nice way to approach the problem. Can you find a simple relationship...

Yes, you are right. The textbook solution is wrong where it starts off with ##F_{net} = i_{ind}LBx_2 - i_{ind}LBx_1##. As you noted, it should be ##F_{net} = i_{ind}LB(x_2) - i_{ind}LB(x_1)##, where ##B(x_1)## is the magnetic field at ##x = x_1## and ##B(x_2)## is the magnetic field at ##x...

The answer should be proportional to ##n^2##, not ##n^3##. Show the details of your calculation so we can help you identify any mistakes. The answer that was provided to you has some typographical errors, but the ##n^2## is correct. [EDIT: Nevermind, I was thinking of finding the current in...

Looks good. Check the last equation above. In the parentheses on the left side, the first term should not have ##B##. It should be ##\dfrac{1}{\rho \epsilon_0}##. This will change your expression for ##\alpha## where you have It should be $$\displaystyle \alpha =\frac{1}{\rho \epsilon_0}...

I think that's right. Of course, for the charge accumulation to increase as the rod accelerates, there must be a small current. So, at any time, qvB must be a little bit greater than E.

My right hand is telling me that the forces on the faces are in the direction to cause the faces to repel each other without any torque on the disk.

Yes. There will not be any current in this case since the magnetic force on a charge carrier is canceled by the electric force from the charge accumulation at the ends of the rod. However the current will be very small since the magnetic and electric forces on a charge carrier in the rod will...

The forces in the joints are internal forces of the system. At a joint where two bars meet, the forces that the two bars exert on each other are action-reaction forces.

OK. Thank you. I agree with you. The internal electric field ##E_i## needs to be taken into account. ##E_i = \sigma/\varepsilon_0## and ##J = \dot \sigma##. I believe you can set up two coupled, first-order differential equations for ##\sigma(t)## and ##v(t)##, where ##v(t)## is the speed...

@AmanWithoutAscarf Can you give us the complete statement of the problem as originally given to you?

I'm not sure I follow this. From the viewpoint of the Earth frame, there is no induced electric field at any point of space since the magnetic field is time-independent in this frame. In this frame, the torque on the ring is due to the tangential component of the magnetic Lorentz force...

Yes, the time derivative of the magnetic flux ##\Phi## through the falling ring will generally be time-dependent.

It's unlikely that a particular video or book suggestion will happen to hit the "sweet spot" for you. This video might be too elementary. It builds up to Einstein's equation with interesting historical commentary. It also discusses the meaning of the symbols in the equation, but not in much...

In the reference frame of the earth, the magnetic field is static. So, there is no induced electric field in this frame. The torque on the loop is due to the Lorentz force associated with the z-component of velocity of the ring and the radial (horizontal) component of the B field. In a...

Your comments look good to me. Normal forces and tension forces are associated with deformations (strains) of materials. They “adjust” to the situation.