- #1
exitwound
- 292
- 1
Homework Statement
Homework Equations
[tex]U = -\vec \mu \cdot \vec B[/tex]
The Attempt at a Solution
As you can see, I calculated [itex]\mu[/itex] = 4.08x10-3 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}{|\mu||B|} = 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:
Code:
μx = 2.45e-3 Bx = .25
μy = -3.27e-3 By = 0
μz = 0 Bz = .3
[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)?