Calculate the net force on square loop (Magnetism)

In summary: To figure out the torque, you could start with the equation for torque using a single loop: \tau = NIABsin\Theta.Multiplying this equation by the length of the wire (in meters) and then dividing by 2 gives you the torque (in newton meter) for the wire.
  • #1
theunbeatable
11
0

Homework Statement



A square loop of wire of side length L lies in the xy-plane, with its center at the origin and its sides parallel to the x- and y- axes. It carries a current i, in a counterclockwise direction, as viewed looking down the z-axis from the positive direction. The loop is in a magnetic field given by B = (B0/a)(zx-hat + xz-hat), where B0 is a constant field strength, a is a constant with the dimension of length, and x-hat and z-hat are unit vectors in the positive x-direction and positive z-direction. Calculate the net force on the loop.

Homework Equations



F = iL x B

The Attempt at a Solution



What's throwing me off is the zx-hat + xz-hat. I'm not sure how to figure this out. When I tried to solve the problem, though, everything canceled out because the current on opposite sides of the square is flowing in opposite directions, so I got the force to be 0. However, I know this answer is not right. Any help would be appreciated.
 
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  • #2
B = (B0/a)(zx-hat + xz-hat)

You mean: $$\vec{B}=\frac{B_0}{a}\left ( z\hat{x} + x\hat{z} \right )$$ i.e. ##\hat{x}## is a unit vector pointing in the positive x direction ... which would more commonly be written: ##\hat{\imath}##.

When I tried to solve the problem, though, everything canceled out because the current on opposite sides of the square is flowing in opposite directions, so I got the force to be 0.
The current flowing in opposite directions may produce the same force in opposite directions - but the forces act at different places, so they do not cancel. Hint: "torque".

The equation you gave for the magnetic field, though, suggests that the field strength, and thus the force for the same current, is different in different places ... so I'd expect the force to be different along different bits of the loop.
 
  • #3
This is what I have so far now. For torque, I have the equation [itex]\tau[/itex] = NiABsin[itex]\Theta[/itex]. Since there is only one loop, N would be 1. And since [itex]\tau[/itex] = F x r, I can equate the two expressions and solve for F.

But I'm still confused with what to do. Do I have to find the torque for each side of the square and then add them to get the net torque?
 
  • #4
theunbeatable said:
This is what I have so far now. For torque, I have the equation [itex]\tau = NiAB\sin\theta[/itex].
This expression is for a specific set of circumstances - do they apply in this situation?

But I'm still confused with what to do. Do I have to find the torque for each side of the square and then add them to get the net torque?
If I were you, I'd start by understanding the first equation you wrote down (it's telling you about the net torque on the loop in terms of the flux enclosed by it) ... or you can go back to first principles with ##\vec{F}=q\vec{v}\times\vec{B}## and work out the force on each bit of the wire.
The fact that it is a square loop makes the geometry easier.

What I need you to do is confirm that what I wrote for the ##\vec{B}## expression in post #2 is correct. You have to actually say - don't make me guess.
 
  • #5
Simon Bridge said:
What I need you to do is confirm that what I wrote for the ##\vec{B}## expression in post #2 is correct. You have to actually say - don't make me guess.

Oh sorry, that was the right equation. I'm bad at formatting so that's why I said x-hat and z-hat.

And it's the magnetic field equation that's really throwing me off. I don't understand what xz-hat and zx-hat (again, sorry with the formatting but I can't seem to find the proper unit vector symbols), are supposed to represent. Are they cross products or planes or something else?

And I'm trying to figure out what to do with torque, but besides the equation I mentioned above, which was derived in my textbook for square loops, I don't have anything else about relating torques to magnetism.
 
  • #6
The "hat" notation indicates a unit vector ... so ##\hat{x}=\hat{\imath}##, ##\hat{y}=\hat{\jmath}##, and ##\hat{z}=\hat{k}## ... you know, like i,j,k notation?
The numbers right in front are the components of the vector.

So: $$\vec{B}=\frac{B_0}{a}\begin{pmatrix} z\\0\\x \end{pmatrix}$$

Presumably you have some relation telling you the force on a wire carrying a current?
You also have ##\vec{F}=q\vec{v}\times \vec{B}##

You have the book's derivation for the formula you wrote too - that should tell you how to derive an equation for this particular situation.
 

What is a square loop?

A square loop is a closed circuit of wire in the shape of a square, typically made of conducting material such as copper. It is used in experiments and calculations involving magnetism.

How is the net force on a square loop calculated?

The net force on a square loop can be calculated by using the equation F=IlBsinθ, where F is the force in Newtons, I is the current in amperes, l is the length of the wire in meters, B is the magnetic field strength in Tesla, and θ is the angle between the direction of the current and the direction of the magnetic field.

What factors affect the net force on a square loop?

The net force on a square loop is affected by the strength of the magnetic field, the current flowing through the loop, the length of the wire, and the angle between the direction of the current and the direction of the magnetic field. Additionally, the material and thickness of the wire can also have an impact on the net force.

What is the direction of the net force on a square loop?

The direction of the net force on a square loop can be determined using the right-hand rule. If you point your thumb in the direction of the current and your fingers in the direction of the magnetic field, the direction of the net force will be perpendicular to both, following the direction of your palm.

How is the net force on a square loop affected by changing the angle between the current and the magnetic field?

The net force on a square loop is directly proportional to the sine of the angle between the current and the magnetic field. This means that as the angle increases, the net force will also increase. When the angle is 90 degrees, the net force will be at its maximum.

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