Calculating an expression for work in a magnetic field

In summary, the conversation involved deriving a formula to find work, which was done by using the formulas W = F * D, V = BA/T, and I = V/R. There was some confusion about whether the formula F = B * (BA/tR) * A was correct, as it seemed to be specific to solenoids. However, it was later determined that the formula was correct and could be used to find the work in this situation. The correct formula for work was found to be (BA)^2 / tR.
  • #1
QuickSkope
60
0
As shown, I must derive a formula to find work.

http://i.imgur.com/0U0eLqm.jpg

I proceeded as follows.

W= F * D
= (B * I * L) * D

V = BA/T

I = V/R
= (BA/t)/R
= BA/tR

F = B * (BA/tR) * A
W = ((B^2*A^2)/tR) * A
W = (B^2 * A^3)/tR

Does that seem right? IIRC, the B*I*L formula is only for solonoids?

Thanks
 
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  • #2
I don't understand this

"F = B * (BA/tR) * A
W = ((B^2*A^2)/tR) * A
W = (B^2 * A^3)/tR

However consider Power = VI and work = power X time
I think you will get a similar ansawer
 
  • #3
I solved L in terms of B, A, R and T, as the question doesn't allow me to use I. Only thing I'm confused about is if I can use V = B * I * L in this situation, as I thought that was for solonoids only.
 
  • #4
What is L? I recall that Force = B X current X length of wire.
Think about it this way. V = dphi/dt and this is what you have.
Power = V^2/r

Work = power X Time, yes?
 
  • #5
barryj said:
What is L? I recall that Force = B X current X length of wire.
Think about it this way. V = dphi/dt and this is what you have.
Power = V^2/r

Work = power X Time, yes?

((BA/t)^2 / r) * t gives me

(BA)^2 / tR, which is a different equation then the other one I got. So is mine wrong?
 
  • #6
I got (BA)^2 / tR. In your post #1 you have an extra A. There might be a math error. Do you have the correct answer?
 
  • #7
Unfortuantetly, I don't have an answer. I'll check it today. Thanks :)
 

1. How is work defined in a magnetic field?

Work in a magnetic field is defined as the energy required to move a charged particle against the magnetic force.

2. What is the formula for calculating work in a magnetic field?

The formula for calculating work in a magnetic field is W = qVBsinθ, where W is work, q is the charge of the particle, V is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the particle's velocity and the direction of the magnetic field.

3. How does the direction of the magnetic field affect the work done?

The direction of the magnetic field affects the work done by changing the angle between the particle's velocity and the direction of the magnetic field. The work done will be highest when the angle is 90 degrees, and no work will be done when the angle is 0 degrees.

4. What is the unit of work in a magnetic field?

The unit of work in a magnetic field is joules (J), which is also the unit of energy.

5. Can work be negative in a magnetic field?

Yes, work can be negative in a magnetic field if the angle between the particle's velocity and the direction of the magnetic field is greater than 90 degrees. This means that the magnetic force is doing work on the particle, rather than the particle doing work against the magnetic force.

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