What is the role of motional emf in opposing motion?

AI Thread Summary
Motional electromotive force (emf) plays a crucial role in opposing motion when a bar is pushed through a magnetic field. As the bar moves, an induced current flows in a direction that creates a magnetic force opposing the applied force, maintaining steady velocity. The discussion highlights that in steady state, the magnetic force equals the applied force, leading to a clockwise current in the system. It clarifies that introducing additional forces is unnecessary if the system is already in equilibrium. Understanding these dynamics is essential for analyzing electromagnetic systems effectively.
sandy.bridge
Messages
797
Reaction score
1

Homework Statement


Hello all,
If we have a bar that is perpendicular to a magnetic field, and that bar is being pushed to the left via an applied force, all the while a current (due to a voltage source and resistor) runs down through the wire. Therefore, there would be a total force comprised of a magnetic force due to the current and an applied for, both directed to the left.

My question is: since another magnetic force develops to oppose this motion and allow dv/dt=0, it will have to have current go UP through the metal rod, and therefore oppose the initial current that was supplied by the voltage source, correct?

Hopefully that makes sense without the need of a picture. If not, please let me know and I will attempt to draw a picture. Regards, Sandy.
 
Physics news on Phys.org
untitled-4.jpg


Here, I added a picture. F is the applied force acting on the bar. The B field is directed out of the picture, and there is a voltage source and resistor in series with the conducting bars. Friction is neglected. Obviously there will be current that enters the system and thus, there will be a magnetic force acting on the wire directed in the same direction as the force F. Now, another force, say F2, will have to induce a current in the opposite direction in order to obtain constant velocity. F2 will act opposite to F and the magnetic force developed due to the initial current flowing.

The current, once steady state is attained, will be clockwise in this picture.

Does this seem right?
 
No, if the system is in steady state, the magnetic force will oppose the velocity, and therefore the current will flow clockwise as you said. However, you introduced an unnecessary force.

All you would need to do is set F=Fmag for steady state conditions.

Feel free to correct me if I am wrong, considering I am new to this type of theory.
 
Also note that the only time that you would have had the conditions as you had it, is if the question explicitly stated that they wanted to prevent the system from moving, ie equilibirium with the sum of forces equal to zero.

In that case, it would be F + F(magnetic) = F(external)
 
I can't see it. Shouldn't there be a magnetic force along with the applied force F in the same direction, that would have to be overcome by another, but stronger magneticforce in the opposite direction?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top