# Magnetic force

#### Dell

in the diagram is a wire which lies in a magnetic field B, what is the force on the wire if a current of I flows through it and its geometry is comprised of 2 straight ends with a length of L each and a half circle with a radius of R??

http://lh6.ggpht.com/_H4Iz7SmBrbk/SiK5CuRdZXI/AAAAAAAABBI/pqhREiis00U/s720/Untitled.jpg [Broken]

what i did was break it up into 3 parts, the 2 ends and the circular part,,,

for all the parts, F=ILxB, but the angle is 90 dor all parts, so F=ILB

for the ends F=ILB
for the cricle F=I*(pi)*R*B

Ftotal=IB(2L+piR)

but the answer ios somehow meant to be
F=2IB(L+R)
cna someone see how this can be?

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#### Redbelly98

Staff Emeritus
Homework Helper
While the angle between "L" and B is indeed 90°, the force on different parts of the semicircle will be at different angles. You'll need to consider x and y components of the force.

#### Dell

but if B is on the Z axis, would any vector on the xy plain not be 90 degrees to B??

#### Redbelly98

Staff Emeritus
Homework Helper
The force vectors will be at different directions within the xy plane. Yes, they are at 90° to B, but they are not all in the same direction.

Again, consider x and y components.

#### Dell

so would i need to integrate ILB for the ciclular part? how would i do that

#### Dell

would the total x not be eqaul to 0 though since the 1st half will be the negative of the 2nd half(for x) leaving me with the integral for y axis

or do i need to do a double integration, of the line y=-sqrt(1-X^2) from (-R,0) to (R,0)

i realy have no idea how to tackle this?

#### Redbelly98

Staff Emeritus
Homework Helper
Yes, the x-components would cancel, leaving just the y-component of force.

What is the y-component of force for a section of the curve dL? That is what you'll need to integrate for the semicircle section. It might be useful to express things in terms of the angle as one integrates along the semicircle path.

#### Dell

can i say that my total Y force is the sum of all the y forces, each one being F*sin(α) since the force is constant only α changes, so can i integrate by α from 0 to 2pi?
α being the angle between the F vector and the Fx vector??

∫F*sinα*dα

F*-cosα (from 0->2pi) = -2F,
and F=ILB=I*(pi*R)*B

=-2pi*R*LB

F=2ILB-2pi*R*B

F=2IB(L-pi*R)

.... still not right :(

#### Redbelly98

Staff Emeritus
Homework Helper
... so can i integrate by α from 0 to 2pi?
No, that would be the case for a full circular loop. But this is only half of a circle. Try from 0 to π instead.

#### Dell

of course, silly me,,, but that is what i meant, and what i calculated,,,

∫F*sinα*dα

F*-cosα (from 0->pi) = -2F,
and F=ILB=I*(pi*R)*B

=-2pi*R*LB

F=2ILB-2pi*R*B

F=2IB(L-pi*R)

.... still not right :(

#### Redbelly98

Staff Emeritus
Homework Helper
Okay, understood.

∫F*sinα*dα

F*-cosα (from 0->pi) = -2F,
and F=ILB=I*(pi*R)*B

=-2pi*R*LB
Think about this. The sine function is positive between 0 and pi, yet you get a negative value for the integral. Something is wrong with what you did here.

#### Redbelly98

Staff Emeritus
Homework Helper
Looking back at your post #8, I don't think you have set up the integral quite right ... I will post another hint/comment shortly ...

#### Redbelly98

Staff Emeritus
Homework Helper
can i say that my total Y force is the sum of all the y forces, each one being F*sin(α) since the force is constant only α changes, so can i integrate by α from 0 to pi?
α being the angle between the F vector and the Fx vector??

∫F*sinα*dα
We need to think carefully what goes in that integral.

The sinα part is correct, since it represents the y-component of force that we are looking for.

That means the other part of the integral should represent the magnitude of the force on a short segment dL of the semicircle. Let's call this force dF

Since
dF = I dL B
and
dL = R dα
you can write out the integral in terms of I, B, and R. And of course, include the sinα term:

sinα dF = force on semicircle​

Hope that helps.

#### Dell

thanks,, really helped a lot,,, only thing i dont understand is how you converted dL into Rdα ??

Staff Emeritus
Homework Helper

thanks

#### bsodmike

Okay, understood.

Think about this. The sine function is positive between 0 and pi, yet you get a negative value for the integral. Something is wrong with what you did here.
I just came across this. First of all I cannot see how you (Dell) went from,

1)
F*-cosα (from 0->pi) = -2F,
and F=ILB=I*(pi*R)*B

to -2pi*R*LB < = Where did that L come from?

1) if you go from 0 to $\pi$ your answer would have simply been F as $-\cos(\pi)=1$

I tried, by expanding from dF through to da,

$IRB\int_{\pi}^{2\pi}\sin(a)\;da=\left[-\cos(a)\right]^{2\pi}_{\pi}=-1-1=-2$

Hence, F_circ = -2IRB ? How does one proceed from here?

#### Redbelly98

Staff Emeritus
Homework Helper
I just came across this. First of all I cannot see how you (Dell) went from,

1)
F*-cosα (from 0->pi) = -2F,
and F=ILB=I*(pi*R)*B

to -2pi*R*LB < = Where did that L come from?
It should have been an I, we missed that earlier

1) if you go from 0 to $\pi$ your answer would have simply been F as $-\cos(\pi)=1$

I tried, by expanding from dF through to da,

$IRB\int_{\pi}^{2\pi}\sin(a)\;da=\left[-\cos(a)\right]^{2\pi}_{\pi}=-1-1=-2$

Hence, F_circ = -2IRB ? How does one proceed from here?
First, don't worry about the negative sign. So this force is just 2IRB, and we'll worry about in what direction it acts later.

Next you need to find the force on the two straight segments.

#### bsodmike

It should have been an I, we missed that earlier

First, don't worry about the negative sign. So this force is just 2IRB, and we'll worry about in what direction it acts later.

Next you need to find the force on the two straight segments.
Thanks for the quick response redbelly :)

Well that's (as earlier stated) F=ILxB and, from the definition of a cross product,

$F=I\left|L\right|\left|B\right|\sin(\pi)=ILB$

Of course there are two segments so the straight segments total is F_straight = 2ILB.

Therefore, ignoring the sign the total F = 2ILB + 2IRB = 2IB(L+R) which is what Dell was looking for. Now, with regards to the negative sign please do elaborate :)

I guess the negative sign is simply a remnant from integrating sin from $\pi$ to $2\pi$; although it's negative with respect to the x-axis it doesn't really apply in this case. On the other hand if the negative sign did indicate direction of force then my earlier summation of the two may not be correct.

#### Redbelly98

Staff Emeritus
Homework Helper
So far so good.

Now, with regards to the negative sign please do elaborate :)
Okay, now that you know the magnitudes of the forces acting on the different wire sections, it's time to use the right-hand-rule to get the directions each of those forces act in. It'll probably help if you just assume a direction for I, even though that is not indicated in the diagram.

Once you have the directions of the different forces, you can figure out whether their magnitudes must be added or subtracted.

#### bsodmike

So far so good.

Okay, now that you know the magnitudes of the forces acting on the different wire sections, it's time to use the right-hand-rule to get the directions each of those forces act in. It'll probably help if you just assume a direction for I, even though that is not indicated in the diagram.

Once you have the directions of the different forces, you can figure out whether their magnitudes must be added or subtracted.
[STRIKE]'right-hand-rule' - do you mean the right hand grip rule (screw rule?). [/STRIKE] I thought using Flemming's left-hand rule (motor effect) would be more suited here; certainly not Flemming's right-hand rule as that is for electromagnetic induction (generator effect). Thanks for the tip.

edit: Just realised you meant http://en.wikipedia.org/wiki/Magnetic_field#Direction_of_force". I'm just used to using Flemming's left-hand rule!

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