# Homework Help: Force required to pull loop from magnetic field

1. May 12, 2007

### t_n_p

1. The problem statement, all variables and given/known data

Part of a single rectangular loop of wire with dimensions shown is situated inside a region of uniform magnetic field of 0.55T. The total resistance of the loop of 0.230ohms. Calculate the force required to pull the loop from the field (to the right) at a constant velocity of 3.4m/s. Neglect gravity.

http://img508.imageshack.us/img508/8268/asdfdr5.jpg [Broken]

3. The attempt at a solution
Really don't know where to start! I'm not sure which equations are relevant as F=nBIL will not work because I don't know the current!

Last edited by a moderator: May 2, 2017
2. May 12, 2007

### Hootenanny

Staff Emeritus
Consider the induced emf in the loop...

3. May 12, 2007

### t_n_p

I only know to formulas involving EMF.

ε=NBAωsing(ωt) and
ε= -N {ΔΦb/Δt}

neither of which seems to help!

4. May 12, 2007

### Staff: Mentor

Review the usage of the second equation.

5. May 12, 2007

### Hootenanny

Staff Emeritus
This should be helpful . Now, the magentic field is constant, so how else can the flux be changing?

Edit: Dammit jt!

6. May 12, 2007

### t_n_p

ahh, ok
N=1, B=0.55T, ΔA=(0.35*0.75)=0.2625.

I'm unsure as to the value of Δt. My understanding is that it is the value of time for which the coil is fully immersed. The length is 0.75m and it is moving right at 3.4m/s. So is Δt simply 0.75/3.4=0.22 seconds?

7. May 12, 2007

### Hootenanny

Staff Emeritus
Careful;

$$\xi = -B\frac{dA}{dt}$$

Now, for constant height;

$$\xi = -B\cdot0.35\frac{dl}{dt} = -B\cdot0.35\cdot v$$

8. May 12, 2007

### t_n_p

Never was taught that formula. Is there any way to come to the same conclusion using the second of the formulas I provided or is this the only way?

Even a similar example (constant height) I have seen utilises the second formula I provided.

9. May 12, 2007

### Hootenanny

Staff Emeritus
This follows directly from Faraday's law and is not something you should learn. Taking Faraday's law and assuming single coil;

$$\xi = -\frac{d\Phi}{dt} = -\frac{d}{dt}(B\cdot A)$$

Now, since the magnetic field is constant we can take it oit of the differential;

$$\xi = -B\frac{dA}{dt} = -B\frac{d}{dt}(h\cdot l)$$

Now, the height of the coil (h) in the magnetic field doesn't change, it is only the length that changes. Hence;

$$\xi = -B\cdot h\frac{dl}{dt}$$

Now, noting that the velocity (v) is in the same direction as the legnth of the coil (l) we can write;

$$\xi = -B\cdot h\cdot v$$

Do you follow? As I said before, this is not something you should learn, it is simply an application of Faraday's law.

Last edited: May 12, 2007
10. May 12, 2007

### t_n_p

Ok, got you!

Now I've got emf I find I from V/R

So I=(-0.55*0.35*3.4)/0.23
I=-2.845652174A

Then I use F=nBIL
F=1*0.55*-2.85*L

Just double checking, L is the length of the loop of wire correct (i.e. 0.75m)?

11. May 12, 2007

### Hootenanny

Staff Emeritus
Looks good to me
Careful! L is not 0.75m (or L in my notation above). L in this case is the height, recall that F = q(v x B) and F = i(L x B). The force is perpendicular to the current and legnth of the wire. So, for the force we wish to calculate we must take L= h, if that makes sense?

Last edited: May 12, 2007
12. May 12, 2007

### t_n_p

Bit confusing, but the above statement is very helpful!
so then F=1*0.55*-2.845652174*0.35
F=-0.5477880435N

negative newtons, is that possible?:surprised

13. May 12, 2007

### Hootenanny

Staff Emeritus
Indeed it is. Remember you have calculated the force exterted by the magnetic field on the wire. All the negative sign indicates is that the force exterted by the magnetic field on the wire is opposite in direction to the velocity. And thats what would would expect according to Lenz's law.

You could if you wished, calculate the force exerted by the magnetic field on the horizontal portions of the wire. However, note that in these sections of wire the current would be travelling in opposite directions, therefore the two forces would be in the opposite directions and since the lengths are equal the forces would cancel. I hope that makes a little more sense

14. May 12, 2007

### t_n_p

Finally finished!
Thanks so much again!

15. May 12, 2007

### Hootenanny

Staff Emeritus
It was a pleasure I hope I didn't confuse you too much ...

16. Mar 21, 2010

### gb14

This was very helpful. Thanks hoot.