Magnatice field, moving rantangle conducting rod, incline plane

[kant]

Imagina an incline plane of angle @. A rod of length L, is at rest( becasuse you are holding it) at the very top of the incline. A magnatice field is pointing up in your mental picture. Find the velocity v of the rod as a function of time once your let go of the rod. Two things in the problem 1) there is no friction involved in this problem. 2) gravitation field exist. Find the velocity v of the rod as a function of time once your let go of the rod.

 

[marlon]

Same here, show us what you have done so far.

regards
marlon

EDIT : you will need to write down the equations of motion along the incline and perpendicular to the incline. Call those direction x and y, if you want. The forces acting here are gravity and the Lorentz force. Write down their components along each direction...Can you achieve this ?

 

[kant]

Same here, show us what you have done so far.
regards
marlon
EDIT : you will need to write down the equations of motion along the incline and perpendicular to the incline. Call those direction x and y, if you want. The forces acting here are gravity and the Lorentz force. Write down their components along each direction...Can you achieve this ?

I got the answer, but i am curious how other people do the problem. it is not that hard, it is kind of visual. want to compare answers?

 

[Gamma]

What is the answer you got?

 

[kant]

v=MgRsin @/ B( square) L( square)cos@( square)

R is the resistence.

 

[Gamma]

If this is an isolated rod (meaning no closed current loop) only force in the direction of the velocity is gravitational. In that case I don't see how you got 'B' in your final answer.

If I assume that somehow this rod is part of a closed circuit, then induced emf,

$LvB cos(\theta) = IR$

The induced current intern give rise to a force along the slope opposite to the velocity.

$F_B = L^2 B^2 v cos(\theta) / R$

Equation of motion,

$m \frac{dv}{dt} = - L^2 B^2 cos(\theta)/R *v + mg sin \theta$

Upon solving this, I get

$v(t) = \frac{mgR sin(\theta)}{L^2 B^2 cos (\theta)} ( 1- exp( \frac{-L^2 B^2 cos (\theta)*t}{mR})$


Gamma

 

[kant]

If this is an isolated rod (meaning no closed current loop) only force in the direction of the velocity is gravitational. In that case I don't see how you got 'B' in your final answer

Is a close loop. I forgot to put that in. Looking at it from the top view, the rod form one side of the loop. The rod can more up or down the incline.

 

[Gamma]

sounds good then.