Normal/Tangential magnetic flux density problem

[satchmo05]

[PLAIN]http://img232.imageshack.us/img232/7172/60746674.png

Homework Statement

The picture attached to this question shows a magnetic slab with μ_r = 50. A thin conducting film (with μ_r = 1) lies on top of the slab and carries a surface current of 1.0[A/mm] directed out of the page. If magnitude of B₁ = 0.01 [T] and θ₁ = 10^o, find magnitude of B₂ and θ₂.

I apologize in advance for my bad recreation of the image in the text!

Homework Equations

1.) B_1n = B_2n
2.) (1/μ₁)B_1t - (1/μ₂)B_2t = J_s

The Attempt at a Solution

So, for starters, I am assuming μ₁ = μ_o = (4*pi)e^-7. Please let me know if I cannot make this assumption, but I believe I can. Because angles are involved that are not perfectly 90^o, I have to include sines and cosines in the equations above. Using formula #1 --> (0.01)*cos(10^o) = B₂cos(θ₂) = 0.009848. Using formula #2 to find θ₂, I can plug in values to have:
(1/μ_o)0.01*sin(10^o) - (795774.72)B₂sin(θ₂) = 1000[A/m]

Solving for θ₂, I find that the angle is 90^o. However, this is not the answer in the back of the book. The answer for θ₂ in the back of the text is 86.2^o and a magnitude of B₂ to be 0.15[T]. I was also not receiving this value for B₂. I have a feeling that my problem is that I cannot assume that μ₁ = μ_o. Please help me! Thanks for all help in advance!

 

[elect_eng]

Please help me! Thanks for all help in advance!

I believe that you have everything essentially correct with one small, but important, exception.

You need to reverse the sign of the current density in your equation. If you look at the orientation, you have set up your equations as if the current was flowing into the page, rather than out of the page.

 

[Antiphon]

uR=50 so u2 is not equal to u1.

 

[elect_eng]

Actually, I see one more error. I think you multiplied $$\mu_0$$ by 50 squared (in region 2) rather than by just $$\mu_r=50$$ as specified.

 

[satchmo05]

I believe that you have everything essentially correct with one small, but important, exception.

You need to reverse the sign of the current density in your equation. If you look at the orientation, you have set up your equations as if the current was flowing into the page, rather than out of the page.

Elect_eng - this still results in an angle of 90^o.

uR=50 so u2 is not equal to u1.

Antiphon - when did I mention this? Can I assume that u₁ = u_o? That was one of my original questions.

Actually, I see one more error. I think you multiplied $$\mu_0$$ by 50 squared (in region 2) rather than by just $$\mu_r=50$$ as specified.

Elect_eng - I carefully went back over my implementation and I still receive 90^o

 

[satchmo05]

Here is my implementation:

1.) B_1n = B_2n
2.) (1/μ₁)B_1t - (1/μ₂)B_2t = J_s
B₁cosθ₁ = B₂cosθ₂ = 0.01cos(10^o) = 0.009848
(1/μ_o)B₁sinθ₁ - (1/μ_r)B₂sinθ₂ = -J_s
(1/μ_o)B₁sinθ₁ + J_s = (1/μ_r)B₂sinθ₂
(μ_r /μ_o)B₁sinθ₁ + μ_rJ_s = B₂sinθ₂
119092.41 = B₂sinθ₂

[B₂sinθ₂ / B₂cosθ₂] = (119092.41/0.009848)
tanθ₂ = 12,093,056
θ₂ = tan^-1(12,093,056)
θ₂ = 90^o

 

[elect_eng]

Here is my implementation:

1.) B_1n = B_2n
2.) (1/μ₁)B_1t - (1/μ₂)B_2t = J_s
B₁cosθ₁ = B₂cosθ₂ = 0.01cos(10^o) = 0.009848
(1/μ_o)B₁sinθ₁ - (1/μ_r)B₂sinθ₂ = -J_s
(1/μ_o)B₁sinθ₁ + J_s = (1/μ_r)B₂sinθ₂
(μ_r /μ_o)B₁sinθ₁ + μ_rJ_s = B₂sinθ₂
119092.41 = B₂sinθ₂

[B₂sinθ₂ / B₂cosθ₂] = (119092.41/0.009848)
tanθ₂ = 12,093,056
θ₂ = tan^-1(12,093,056)
θ₂ = 90^o

The permeability in region 2 is $$\mu_0 \mu_r$$ not $$\mu_r$$.

 

[satchmo05]

That was the problem. Thanks for your help and persistence Elect_Eng!