Electromagnetism Q: Physics A level AQA Watch

jl2
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Looking at the diagram B was my first, obvious choice, but a couple is a pair of forces in the same plane but in opposite directions, then i remembered f = bil, and so deduced the area was irrelevant? thus picked the largest value of l, which is option A (which was wrong). Can anyone explain the theory behind why B is the one that experiences the largest couple, and which equation i should use to calculate force given an area? The only force ones I know are F=BIL and F=Bqv

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jl2
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(Original post by BobbJo)
I think the picture you wanted to attach didn't attach
Damn thanks for pointing that out ^
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Anonymouspsych
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(Original post by jl2)
Looking at the diagram B was my first, obvious choice, but a couple is a pair of forces in the same plane but in opposite directions, then i remembered f = bil, and so deduced the area was irrelevant? thus picked the largest value of l, which is option A (which was wrong). Can anyone explain the theory behind why B is the one that experiences the largest couple, and which equation i should use to calculate force given an area? The only force ones I know are F=BIL and F=Bqv

Thanks
Ok first things first remember the question asks which loop experiences the largest couple, not the largest force.

Now if you remember to calculate a couple you multiple the magnitude of the force on either end by the distance between the ends at which the equal and opposite coplanar forces are applied.

Also remember that the parts of the loops which are parallel to the direction of B will experience no force (i.e. top of all the loops experience 0 force)- so only the sides of the loops experience the force (the direction will depends on the direction of the current via Fleming's LHR)

So now you can work out the couple experienced by all the loops

Couple on loop A = BL * 0.2L = 0.2BL^2
Couple on loop B = 0.5BL * 0.5L = 0.25BL^2
Couple on loop C = 0.5BL * 0.3L = 0.15BL^2
Couple on loop D = 0.8BL * 0.2L = 0.16BL^2

Clearly the answer is B
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jl2
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(Original post by Anonymouspsych)
Ok first things first remember the question asks which loop experiences the largest couple, not the largest force.

Now if you remember to calculate a couple you multiple the magnitude of the force on either end by the distance between the ends at which the equal and opposite coplanar forces are applied.

Also remember that the parts of the loops which are parallel to the direction of B will experience no force (i.e. top of all the loops experience 0 force)- so only the sides of the loops experience the force (the direction will depends on the direction of the current via Fleming's LHR)

So now you can work out the couple experienced by all the loops

Couple on loop A = BL * 0.2L = 0.2BL^2
Couple on loop B = 0.5BL * 0.5L = 0.25BL^2
Couple on loop C = 0.5BL * 0.3L = 0.15BL^2
Couple on loop D = 0.8BL * 0.2L = 0.16BL^2

Clearly the answer is B
Ah I see, couples and torque as opposed to just the force applied. OK area is not important here.

After looking up the torque equation I have a question though:
- As torque = force x distance d from axis
- And F=BIL,
- Torque = BIL(d)
- Thus for each loop the value of d would be half the stated width, as that would be the "entire width" rather than just the "distance from axis"? i.e. for A: d=0.1L, B: d=0.25L etc...
- So if i was asked to calculate the value of the couple on loop B would it be correct to say:
- t = Fd, F=BIL
- t=BILd = B(1)(0.25L)^2 as opposed to B(1)(0.5L)^2 in your answer?

tytyty
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jl2
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(Original post by BobbJo)
Force on left vertical limb is equal and opposite to force on right vertical limb. This couple causes a torque given by

T=BIxy
x and y being the linear width and length

I = 1A so we get
T=Bxy

hence we calculate torques
A: 0.2BL^2
B: 0.25BL^2
C: 0.15BL^2
D: 0.16BL^2

clearly it is B
t=Fd, F=BIL
t=BILd (L and d are length from axis), so from your equation of t=Blxy, is the y also supposed to be width? How does the length (assumed to be the vertical value) relate to the torque?
thanks
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jl2
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Oh ok that clears it up : )
One more thing, do we still use the entire width in T=BIxy though? Or do we use half the width as say for example A the distance from the axis the loop is spinning around would be 0.1L as opposed to 0.2L?
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jl2
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(Original post by BobbJo)
Yes I adressed this in the post #7

One way of calculating the net moment (torque)

Taking moments about the left vertical limb of the loop,
We have a net moment given by
Moment = Force x distance = BI x length x perpendicular distance of line of action of force from point = (BI x length) x width = BI x length x width
bracketed part is the force for clarity

Another way

Take moments about centre, we have a net moment given by
Moment = Sum of moments of 2 forces = BI x length x perpendicular distance of line of action of first force from point + BI x length x perpendicular distance of line of action of first force from point
= BI x length x (width/2) + BI x length x (width/2)
= BI x length x width

I avoided any ambiguous notation involving L since the symbol L is used in the question for both length and width.

I tried my best
Oh completely my bad, thank you so much. I didnt see post 7 for some reason. Thank you for all the help man : )
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Eimmanuel
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(Original post by jl2)
Ah I see, couples and torque as opposed to just the force applied. OK area is not important here.

....
Are you really sure area is NOT important in the calculation of the couple of moment here?
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jl2
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(Original post by Eimmanuel)
Are you really sure area is NOT important in the calculation of the couple of moment here?
hm, would you like to point it out? Is it not only length from axis that is relevant for moments questions?
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Eimmanuel
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(Original post by jl2)
hm, would you like to point it out? Is it not only length from axis that is relevant for moments questions?
Let say the length of wire that is perpendicular to the magnetic field is  \ell .

The magnetic force is exerting on the length of wire is

 F=BI\ell

As for the couple,

  \tau =BI\ell \times w

where w is the width of the loop of wire.

Note that  \ell \times w is the area of the loop, so


  \tau =BIA
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jl2
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(Original post by Eimmanuel)
Let say the length of wire that is perpendicular to the magnetic field is  \ell .

The magnetic force is exerting on the length of wire is

 F=BI\ell

As for the couple,

  \tau =BI\ell \times w

where w is the width of the loop of wire.

Note that  \ell \times w is the area of the loop, so


  \tau =BIA
Thank you, this helped me in my test ^^
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