Register to reply 
Magnetic Potential Energy 
Share this thread: 
#1
Oct2709, 08:39 PM

P: 292

1. The problem statement, all variables and given/known data
2. Relevant equations [tex]U = \vec \mu \cdot \vec B[/tex] 3. The attempt at a solution As you can see, I calculated [itex]\mu[/itex] = 4.08x103 and got the torque on the loop which is shown in the answer above. The potential energy is defined above as [tex]U = \vec \mu \cdot \vec B = \mu B\cos\theta[/tex] I have to find the angle between the two vectors which I can do using: [tex]\frac{\vec \mu \cdot \vec B}{\muB} = cos \theta[/tex] But what I came up with wasn't right. From the first part, [tex]\mu \vec n = (4.08x10^3)(.6\hat i .8 \hat j) = 2.45x10^3 \hat i  3.27x10^3 \hat j[/tex] If:
[itex]\mu^2 = (2.45x10^3)^2 + (3.27x10^3)^2[/itex] [itex]\mu = 4.09x10^3 [/itex] [itex]B^2 = (.25)^2 + (.3)^2[/itex] [itex]B = 3.91x10^1[/itex] [tex]\vec \mu \cdot \vec B = \mu_xB_x + \mu_yB_y + \mu_zB_z[/tex] [tex]\frac{\mu_xB_x + \mu_yB_y + \mu_zB_z}{\mu B} = cos \theta[/tex] When I put in the numbers, I get 66.8 degrees between the two vectors. Putting this back into the [itex]U = \vec \mu \cdot \vec B = \mu B\cos\theta[/itex] equation gives the wrong answer. Where's the mistake(s)? 


#2
Oct2709, 08:52 PM

HW Helper
P: 4,433

W = F.d = Fx*dx + Fy*dy + Fz*dz. No need to find the angle.



#3
Oct2709, 08:59 PM

P: 292

What's Dx Dy and Dz??



#4
Oct2709, 10:27 PM

P: 151

Magnetic Potential Energy
Those are partial derivatives. To take a partial derivative with respect to x (dx), you hold the y and z terms constant and then take the derivative with respect to x as if it were the only variable.
For dx + dy + dz you do this for all three variables and then add up all the derivatives you found. This tells you the slope in all three directions, which is necessary for a threedimensional problem. He's saying that you can calculate the potential energy of the field by considering the force that an imaginary particle feels in the field at that point. To do this, you need to understand the force that the particle feels in each of the three directions. The force in the x direction times dx, the force in the y direction times dy, and the force in the z direction times dz. If you dot these together, you get the total potential energy. 


#5
Oct2709, 10:30 PM

P: 292

I know what his response meant, but I don't know what dx is. What am I measuring "distance" of? What's the d in F dot d?



#6
Oct2709, 10:36 PM

P: 151

Pardon me for trying to help.
The d is arbitrary because potential energy is not an absolute quantity, but is relative to a chosen origin. For example, the gravitational potential energy at the surface of the earth is not really zero since gravity is still pulling on objects there. However, in physics it is convenient to call that point zero so that you can consider how much more potential energy there is at a height h. In reality, there is no absolute definition for the potential at the surface of the Earth, so an origin must be chosen. In problems where an origin is not explicitly defined the convention is to take the potential energy to be zero at a distance of infinity from the point of interest. Then you can think of the potential energy of the field to be the amount of work required to bring the test particle from infinity to the point of interest through the field. Good luck. 


#7
Oct2709, 10:42 PM

P: 292

Indeed. Didn't mean to come off as harsh. The internet can do that sometimes.
I think his first response has generated even more confusion than there was at the start due to its brevity. I appreciate the help. I just don't know how to complete the problem, obviously. 


#8
Oct2709, 11:04 PM

HW Helper
P: 4,433

Sorry for the brevity.
I have given the definition of the work, to avoid the direct solution. In the given problem μ = μx*i + μy*j +μz*k = ....? B = Bx*i + By*j + Bz*k =........? Now work done is w =  μ.B =  (μx*Bx + μy*By + μz*Bz) = .....? 


#9
Oct2809, 03:23 PM

P: 292

okay that worked. i'm still hazy on work in general. I never could understand it.



#10
Oct2909, 12:19 AM

P: 151

Just think of work as the amount of energy necessary to apply a force for a certain amount of time. If you want to throw a ball 5 feet, you need a certain amount of energy (or work). To throw the ball 25 feet, you need five times as much.
After you have intuition about work as a force that you can relate to in daily life, you can apply that intuition to forces that you don't experience daily (like magnetic potentials). Start from the known to understand the unknown. 


Register to reply 
Related Discussions  
Potential energy for magnetic fields  Introductory Physics Homework  2  
Can magnetic dipoles have potential energy?  Classical Physics  1  
Potential Energy for magnetic dipole  Introductory Physics Homework  1  
Changing Potential Energy of a Magnetic Coil  Introductory Physics Homework  3  
Electric potential, potential difference, and potential energy  Introductory Physics Homework  2 