# Force acting on a wire

## Homework Statement

A 15 turn rectangular metal loop has a resistance of 10 Ohms and dimensions x = 5 cm and y = 10 cm. The loop moves to the right with a constant speed of 4.0 m/s (Some external machine keeps the speed constant despite any other forces.) The loop enters a region from the left with constant magnetic field B = 150mT out of the page.

(i) What is the force (direction and magnitude) acting on the wire when the wire is entering the magnetic field?

F = BIL
I = EMF/R
EMF = vBL
Lenz's Law

## The Attempt at a Solution

I think that the direction of the force is going right to left opposing the force of the machine because that is what Lenz's law states (Maybe). However to find the force I'm unsure if I have the correct equation F = BIL and if i do I'm not sure which dimension to use as the L the x or the y. And I also need to use L to find I in the first place because I = EMF/R.
You can get around this by using an alternate equation for EMF:

EMF = -N$$\frac{\Delta \Phi}{\Delta t}$$

Where I use the Area of the loop instead of just a length (To find the flux), but I don't have a change in time to work with. Maybe I can use the velocity to get that somehow?

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collinsmark
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Where I use the Area of the loop instead of just a length (To find the flux), but I don't have a change in time to work with. Maybe I can use the velocity to get that somehow?
Yes, the velocity is important.

(a) Flux is a function of area.
(b) As the loop enters the magnetic field, the side of the loop within the field sweeps out an area per unit time. In other words, when the loop enters the field, the relevant area is a function of time.
(c) The rate of the change of the swept out area with respect to time is a function of the loop's velocity.

Hmmm. So would it be correct for me to say I = EMF/R Where EMF = NBLv ?

Lv is essentially the area being swept out per unit of time as you said.
If I do this then I get 15(.150T)(.10m)(4m/s)/10$$\Omega$$ = .09A

If that is correct then I think the force acting on the wire entering the field is then F = BIL
= (.150T)(.09A)(.10m) = 1.35 x 10-3 N to the left (I'm still confused on the L, not sure if I'm using the correct value)

collinsmark
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Hmmm. So would it be correct for me to say I = EMF/R Where EMF = NBLv ?

Lv is essentially the area being swept out per unit of time as you said.
If I do this then I get 15(.150T)(.10m)(4m/s)/10$$\Omega$$ = .09A
I think that's reasonable, assuming the 10 cm side of the wire is the side that is perpendicular to the motion; i.e. the loop is moving along the x-axis in the positive direction (it's hard to tell without a figure).

If that is correct then I think the force acting on the wire entering the field is then F = BIL
= (.150T)(.09A)(.10m) = 1.35 x 10-3 N to the left (I'm still confused on the L, not sure if I'm using the correct value)
You may have forgotten about the N part. Remember, it's as if you have 15 wires each contributing a force of F = BIL.

I think that's reasonable, assuming the 10 cm side of the wire is the side that is perpendicular to the motion; i.e. the loop is moving along the x-axis in the positive direction (it's hard to tell without a figure).

You may have forgotten about the N part. Remember, it's as if you have 15 wires each contributing a force of F = BIL.
Yes the 10 cm side is perpendicular to the motion in the figure.

I didn't know that the number of wires had an effect on the force actually, but it would make sense. So if i just multiply this by 15 i'll get the correct force then?

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collinsmark
Homework Helper
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Yes the 10 cm side is perpendicular to the motion in the figure.

I didn't know that the number of wires had an effect on the force actually, but it would make sense.
Think of each turn of wire as a separate wire. Each wire is carrying 0.09 A of current. Thus each wire would contribute F = BIL
= (.150T)(.09A)(.10m) = 1.35 x 10-3 N of force. So you need to add up the forces created by each wire to get the total force.

So if i just multiply this by 15 i'll get the correct force then?
That's what I would do. 