# Friction with induced coil and magnetic force

1. Jun 8, 2012

### IIK*JII

1. The problem statement, all variables and given/known data
As shown in the attached figure, two rails of conducting wire,M and N, run parallel to each other on a horizontal plane and are separated by distance l. The rails are in a magnetic field of magnetic flux density B that is in the vertically upward direction. An electrically conductive bar of mass m is placed on the rails, perpendicular to their length. When a battery of electromotive force E and negligible internal resistance is connected to the rails, the bar begins to slide along the top of the rails. The coefficient of kinetic friction between the bar and the rails is μ'. The part of the bar between the rails has electrical resistance R. The electrical resistance of the rails is negligible.

The bar undergoes linear uniform motion with a speed of v, determine v.

2. Relevant equations
E = $\frac{I}{R}$ ...(1)
E = $\frac{ΔBA}{Δt}$ = Blv ...(2)
ƩFx=0 (bar undergoes linear uniform motion with a speed of v) ...(3)
FB=IlB ...(4)
f = μ'N ...(5)
ƩFy=0 ..(6)

3. The attempt at a solution

I wrote the FBD as seen in attached figure, thus (3) becomes
ƩFx=0
FB=f
substitute (4) and (5); IlB=μ'N .... (7)
substitute (1) for I and from (6) I know N=mg

So, (7) is ElB = μ'mgR ..(8)
Using (2); B2l2v=μ'mgR
∴v =$\frac{μ'mg}{(Bl)^2}$

But I think I miss something. Does anybody tell me what?? I try to find it because my answer was wrong

I appreciate all kinds of any help :!!)
Thankss

File size:
12.4 KB
Views:
53
2. Jun 8, 2012

### truesearch

I think I got the same !!!! what makes you think you got it wrong?... what answer have you been given?

Last edited: Jun 8, 2012
3. Jun 8, 2012

### IIK*JII

The answer is v = $\frac{EBl-μ'mgR}{(Bl)^2}$,, so I think I miss something

4. Jun 9, 2012

### truesearch

I think I have it.
Force on the wire = BIL = μmg (moving at constant velocity)

The moving wire generates an induced emf = e and I = (E-e)/R

e = BLv
If you make all the substitutions you get their answer.
Hope you can work it through.

5. Jun 9, 2012

### IIK*JII

Thank you truesearch

I can work through it,,

Now I knew that I miss the term of induced emf by moving wire

In next problem I will consider more and more carefully

Thanks a lot

IIK*JII

6. Jun 9, 2012

### truesearch

I did the same! Forgot to include the emf induced by the moving wire......it is called the "back emf"......good question !