Velocity of a bar on parallel conducting rails

In summary, the bar moves to the right with a constant force applied and the velocity is a function of R and t.
  • #1
abear247
4
0

Homework Statement



A pair of parallel conducting rails a distance l apart in a uniform magnetic field B⃗ . A resistance R is connected across the rails. The bar has mass m and is initially at rest. A constant force F⃗ e to the right is applied to the bar. Find its velocity as a function of time in terms of R. l, B, m, Fe, and t

Homework Equations



F=ma

The Attempt at a Solution



First part of the question asked to formulate Newton's second law for the bar, and my answer was m(dv/dt) = Fe - (B^2*l^2*v)/R.

This is correct, but trying to get v(t), i move Fe to one side, everything else to the other, then factor v out (not sure if that is allowed...) and then divide the Fe by what is left. I get:

(Fe*R*t)/(m*R+B^2*l^2*t)

Any help would be greatly appreciated, thanks.
 
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  • #2
Hello Bear and welcome to PF,

Good thing you know your first part is OK. I agree.

Next, bringing Fe to one side gets you an equation with v and dv/dt on the righthand side, a real differential equation ! v and dv/dt are different beasts and how you factor them out is a mystery to me.

Any experience with differential equations ?
 
  • #3
Thanks for the reply, i was way too tired yesterday and not thinking very well... And i have some, but i really struggle with them (been awhile since i took calculus, and i only took the first half). If Fe is on one side, is the next step to evaluate m(dv/dt) from vo to v, and since it is initially at rest it leaves just mv? then for (B^2*l^2*v)/r from 0 to v(t) gives (-2B2lv^2)/r

This still does not seem correct to me though...
 
  • #4
No, to me neither. You can check if what is left over satisfies the differential equation ...which was
m(dv/dt) = Fe - (B^2*l^2*v)/R. Of the form $$ {dv\over dt} + \alpha \, v =\beta, \quad \quad \alpha = {B^2\,l^2\over mR},\quad \beta = {F_e\over m} $$ give it a go, good to freshen up calc!
 
  • #5
Would the answer to that be v(t) = β/(1 + [itex]\alpha[/itex]t)?
 
  • #6
Well, what do you get when you fill it in in the differential equation ?
$$ {dv\over dt} + \alpha \, v = {-\alpha \beta \over (1+\alpha t)^2} \ + {\alpha \beta \over 1 + \alpha t} $$ which doesn't look like ## \beta ## to me.

Did you ever get to solve something like $$ {dv\over dt} + \alpha \, v = 0 $$in the past ?
 
  • #7
Hey, well the assignment is over now, the correct answer was (Fe*R/B^2*l^2)(1-e^(-b^2*l^2*t/Rm)

My final attempt was Fe/m - e^(B^2*l^2*t/Rm)

I still got an 88 on the assignment, so its not the worst, thanks for the help though
 

Related to Velocity of a bar on parallel conducting rails

1. What is the velocity of a bar on parallel conducting rails?

The velocity of a bar on parallel conducting rails is determined by the strength of the magnetic field, the length of the bar, and the angle at which it is moving in relation to the magnetic field. It can be calculated using the formula v = Bℓsinθ, where v is the velocity, B is the strength of the magnetic field, ℓ is the length of the bar, and θ is the angle of motion.

2. How does the velocity of a bar on parallel conducting rails change with varying magnetic field strength?

The velocity of a bar on parallel conducting rails is directly proportional to the strength of the magnetic field. This means that as the magnetic field strength increases, the velocity of the bar will also increase. Conversely, as the magnetic field strength decreases, the velocity of the bar will also decrease.

3. What factors affect the velocity of a bar on parallel conducting rails?

The velocity of a bar on parallel conducting rails is affected by the strength of the magnetic field, the length of the bar, and the angle at which it is moving in relation to the magnetic field. Additionally, the material of the bar and the resistance of the rails can also have an impact on the velocity.

4. Can the velocity of a bar on parallel conducting rails ever be negative?

No, the velocity of a bar on parallel conducting rails cannot be negative. This is because the velocity is a vector quantity, meaning it has both magnitude and direction. Even if the bar is moving in the opposite direction of the magnetic field, the velocity will still be positive.

5. How does the angle of motion affect the velocity of a bar on parallel conducting rails?

The velocity of a bar on parallel conducting rails is directly proportional to the angle of motion. This means that as the angle increases, the velocity will also increase. However, if the bar is moving parallel to the magnetic field (θ = 0), the velocity will be 0 regardless of the strength of the magnetic field or the length of the bar.

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