# A problem on magnetic field

#### Kudo Shinichi

1. Homework Statement
A current i flows through a triangular loop in the y-z plane. The loop is situated in a uniform magnetic field B.
http://s5.tinypic.com/14csew3.jpg
a) Find the total force on the loop when the magnetic field is in i) the x-direction, and ii) the y-direction
b) What is the magnetic dipole moment of the loop? (use mu=i x area and the appropriate direction)
c) What is the torque on the loop and how will it tend to turn when the field is in i) the x-direction, and ii) the y-direction?

3. The Attempt at a Solution
a)x-direction:
F=current x length x uniform magnetic field x sinθ
F=-i x B x l
y-direction:
F=i x B x l cos 45

b) area of a triangle: 1/2(b*h)
since the angle is 45 degree so both height and base are the same.
1/2(l^2)
mu=i*1/2(l^2)
however, I don't really know how to figure out the direction for this question.

c) torque=number of loops*current*area*uniform magnetic field
x-direction:
torque=1*i*1/2(l^2)*B
y-direction:
torque=1*i*cos45*1/2(l^2)*B

I am not really sure whether I did the problem correctly or not...and for some questions I only know how to do part of them

Any comment or help would be great! Thank you very much.

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#### Redbelly98

Staff Emeritus
Homework Helper
1. Homework Statement
A current i flows through a triangular loop in the y-z plane. The loop is situated in a uniform magnetic field B.
http://s5.tinypic.com/14csew3.jpg
a) Find the total force on the loop when the magnetic field is in i) the x-direction, and ii) the y-direction
b) What is the magnetic dipole moment of the loop? (use mu=i x area and the appropriate direction)
c) What is the torque on the loop and how will it tend to turn when the field is in i) the x-direction, and ii) the y-direction?

3. The Attempt at a Solution
a)x-direction:
F=current x length x uniform magnetic field x sinθ
F=-i x B x l
This looks like the force on one of the short segments of the loop. You'll need to add up the forces for each segment, in order to get the total force on the loop.

y-direction:
F=i x B x l cos 45
Same applies, you'll need to get the force on each segment and add them up.

b) area of a triangle: 1/2(b*h)
since the angle is 45 degree so both height and base are the same.
1/2(l^2)
mu=i*1/2(l^2)
however, I don't really know how to figure out the direction for this question.
Use the right-hand rule. Curl your fingers in the direction of current, and your thumb points in the direction of the magnetic moment.

c) torque=number of loops*current*area*uniform magnetic field
x-direction:
torque=1*i*1/2(l^2)*B
y-direction:
torque=1*i*cos45*1/2(l^2)*B
That doesn't seem right. Again, forces on each segment. Treat each force as acting at the center of the segment, and see how they combine to produce a net torque.

#### Kudo Shinichi

This looks like the force on one of the short segments of the loop. You'll need to add up the forces for each segment, in order to get the total force on the loop.

Same applies, you'll need to get the force on each segment and add them up.

Use the right-hand rule. Curl your fingers in the direction of current, and your thumb points in the direction of the magnetic moment.

That doesn't seem right. Again, forces on each segment. Treat each force as acting at the center of the segment, and see how they combine to produce a net torque.
a)F=current x length x uniform magnetic field x sinθ
x-direction:
F=(-i x B x sqrt(l^2+l^2))+(i x B x l x sin 45) + (i x B x cos 45)
y-direction:
F=(-i x B x l)+(i x B x l x cos45)+(i x B x l x tan45)

b) direction of magnetic dipole moment: 45 degree respect to the horizontal negative y axis or 135 degree

I still don’t really understand how to solve part c, isn’t torque on a loop equals to magnetic dipole moment times the uniform magnetic field? Can you expalin more on this part? thank you very much

#### Redbelly98

Staff Emeritus
Homework Helper
a)F=current x length x uniform magnetic field x sinθ
x-direction:
F=(-i x B x sqrt(l^2+l^2))+(i x B x l x sin 45) + (i x B x cos 45)
There seems to be a couple of problems still (or maybe I don't understand what you are doing?):
1. Each segment is at 90 degrees to the direction of B (+x direction), so I don't see where these sin(45) and cos(45) terms are coming from.
2. Forces are vectors, so use vector addition to add them up.

y-direction:
F=(-i x B x l)+(i x B x l x cos45)+(i x B x l x tan45)
Again, I'm rather confused by how you got what you did. So just let me work an example out for you:

For the vertical wire, current (-z direction) is at 90 degrees to B (+y direction). So the force's magnitude is
F1 = i B l sin(90) = i B l
And, by the right-hand rule, F1 is in the +x direction.

You'll need to do that for each of the 3 sides, then add the forces as vectors. (And keep in mind that each side is at a different angle to B.)

b) direction of magnetic dipole moment: 45 degree respect to the horizontal negative y axis or 135 degree
Actually, the dipole moment is always perpendicular to the plane of the loop. So that will be either the +x or -x direction.

I still don’t really understand how to solve part c, isn’t torque on a loop equals to magnetic dipole moment times the uniform magnetic field? Can you expalin more on this part? thank you very much
Ah, you're right (Sorry, I had to go back and review this).
More accurately, torque is the cross product μ x B, where μ is the magnetic dipole moment.

#### Kudo Shinichi

There seems to be a couple of problems still (or maybe I don't understand what you are doing?):
1. Each segment is at 90 degrees to the direction of B (+x direction), so I don't see where these sin(45) and cos(45) terms are coming from.
2. Forces are vectors, so use vector addition to add them up.

Again, I'm rather confused by how you got what you did. So just let me work an example out for you:

For the vertical wire, current (-z direction) is at 90 degrees to B (+y direction). So the force's magnitude is
F1 = i B l sin(90) = i B l
And, by the right-hand rule, F1 is in the +x direction.

You'll need to do that for each of the 3 sides, then add the forces as vectors. (And keep in mind that each side is at a different angle to B.)

Actually, the dipole moment is always perpendicular to the plane of the loop. So that will be either the +x or -x direction.

Ah, you're right (Sorry, I had to go back and review this).
More accurately, torque is the cross product μ x B, where μ is the magnetic dipole moment.
I have re-read the textbook again, and I found out that the direction of the force is always perpendicular to the B, and if you use the right-hand rule, fingers point straight along the current and bend to get B, which means that B is also perpendicular to current...so I am kind of confused now, if B is always perpendicular to current, then shouldn’t the angle to B always be the same?
I have tried two ways to solve for both directions, one is follow the right-hand rule, B is always perpendicular to current, and the second way is determine the angles from the B-direction you told me
Set Vertical wire F1, horizontal wire F2, and the hypotenuse of the horizontal and vertical wires F3
x-direction:
The direction of B is always perpendicular to the force
The length for both F1 and F2 are the same
F1=F2=i x B x l sin90= i x B x l
F3: the length of this wire is sqrt(l^2+l^2)
F3=i x B x sqrt(l^2+l^2) sin90=i x B x sqrt(l^2+l^2)
Total force: (i x B x sqrt(l^2+l^2))i+(i x B x l)j+(i x B x l)k
B-direction(+x direction)
F2= i x B x l sin(90+45)
F1= i x B x l sin(180+45)
F3=0 because it is parallel to direction of B
Total force= (i x B x l sin(90+45))j + (i x B x l sin(180+45))k

y-direction:
The direction of B is always perpendicular to the force
F1= i x B x l sin90= i x B x l
F2 = i x B x l sin90 = i x B x l
F3= i x B x sqrt(l^2+l^2) sin90=i x B x sqrt(l^2+l^2)
Total force: (i x B x sqrt(l^2+l^2))i+(i x B x l)j+(i x B x l)k

B-direction(+y direction)
F1= i x B x l sin90= i x B x l
F2=0 because it is parallel to direction of B
F3= i x B x l sin(180+45)
Total force= (i x B x l)k +( i x B x l sin(180+45))i
b) the direction would be –x direction.
c)
If the the direction of B is always perpendicular to the force then torque in both x and y direction should be the same.
x-direction:
torque=i*1/2(l^2)*B
y-direction:
torque=i*1/2(l^2)*B

#### Redbelly98

Staff Emeritus
Homework Helper
I have re-read the textbook again, and I found out that the direction of the force is always perpendicular to the B, and if you use the right-hand rule, fingers point straight along the current and bend to get B, which means that B is also perpendicular to current...so I am kind of confused now, if B is always perpendicular to current, then shouldn’t the angle to B always be the same?
No, it does not mean B is always perpendicular to the current. You don't have to bend your fingers through 90 degrees, just bend them enough to make them point along B.

I have tried two ways to solve for both directions, one is follow the right-hand rule, B is always perpendicular to current, and the second way is determine the angles from the B-direction you told me
Set Vertical wire F1, horizontal wire F2, and the hypotenuse of the horizontal and vertical wires F3
x-direction:
The direction of B is always perpendicular to the force
The length for both F1 and F2 are the same
F1=F2=i x B x l sin90= i x B x l
F3: the length of this wire is sqrt(l^2+l^2)
F3=i x B x sqrt(l^2+l^2) sin90=i x B x sqrt(l^2+l^2)
Total force: (i x B x sqrt(l^2+l^2))i+(i x B x l)j+(i x B x l)k
You have the correct magnitudes for F1, F2, and F3. However, they will all point in different directions. You need to figure out the direction for each individual force, then add them as vectors.

B-direction(+x direction)
F2= i x B x l sin(90+45)
F1= i x B x l sin(180+45)
F3=0 because it is parallel to direction of B
Total force= (i x B x l sin(90+45))j + (i x B x l sin(180+45))k
I'm not following what you are doing here. Since each segment is perpendicular to x (the direction of B), all angles are 90.

y-direction:
The direction of B is always perpendicular to the force

B-direction(+y direction)
F1= i x B x l sin90= i x B x l
F2=0 because it is parallel to direction of B
F3= i x B x l sin(180+45)
What is the length of segment #3?
Total force= (i x B x l)k +( i x B x l sin(180+45))i
Think carefully about the directions of F1 and F3.
Correct that F2=0

b) the direction would be –x direction.
Yes.
c)
If the the direction of B is always perpendicular to the force then torque in both x and y direction should be the same.
x-direction:
torque=i*1/2(l^2)*B
y-direction:
torque=i*1/2(l^2)*B

#### Kudo Shinichi

No, it does not mean B is always perpendicular to the current. You don't have to bend your fingers through 90 degrees, just bend them enough to make them point along B.

You have the correct magnitudes for F1, F2, and F3. However, they will all point in different directions. You need to figure out the direction for each individual force, then add them as vectors.

I'm not following what you are doing here. Since each segment is perpendicular to x (the direction of B), all angles are 90.

What is the length of segment #3?
Total force= (i x B x l)k +( i x B x l sin(180+45))i
Think carefully about the directions of F1 and F3.
Correct that F2=0

Yes.
a) total force in x-direction: sqrt((i x B x l)^2+(i x B x l)^2)+(i x B x sqrt(l^2+l^2))
sqrt((i x B x l)^2+(i x B x l)^2) is the total force of F1 and F2 and the direction is same as F3

total force in y-direction:
since F2 is 0, so we just add up the F1 and F3 which is eqaul to sqrt((i x B x l)^2+(i x B x sqrt(l^2+l^2))^2)

Also I am wondering did I have the correct answer for part c?

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