# Induced tension magnetic field (B)

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1. Mar 6, 2015

### Basip

1. The problem statement, all variables and given/known data
I have the strength of the magnetic field, B, the time, Delta t, a circle formed ring with the diameter, d. I should calculate the induced tension, when the surface is

(a) parallel to the B field
(b) 50 degrees on the B field

Need help to solve it (symbolic if possible).

2. Relevant equations
I thought of something like:
$$U_H = A_H \frac{I B}{d}$$,
but I don't have "I". And there are no angles there. Please tell me how to solve it symbolic.

3. The attempt at a solution

2. Mar 6, 2015

### BvU

Hello Basip, welcome (back) to PF !

Could you please re-read your posting and complete the problem statement ? Perhaps even add a little drawing ? It is now rather unclear what the exercise wants you to do. It's fine if you mention $\Delta t$, but what it its role in this problem ?

Also all and any symbols you want to use in part 3 need clarification. $U_H$ is probably the induced emf ? and $A_H$ an area ? Don't let us guess unnecessarily !

And you can do displayed equations with $$U_H = A_H \frac{I B}{d} $$ to get $$U_H = A_H \frac{I B}{d}$$ and in-line equations with $\#\#$U_H = A_H \frac{I B}{d}$\#\#$ to get things like $U_H = A_H \frac{I B}{d}$

However, your equation doesn't make much sense to me. You sure it fits in the problem context ? Could it be you need something else ?

3. Mar 7, 2015

### Basip

Dear BvU!

Thank you for your help so far. I don't know how to edit my question, so I reply here.

Variables to play with
$$B=0.58T$$
$$\Delta t=0.10s$$
$$d=0.105m$$

Question formulation
In a magnetic field, B, at the time, $\Delta t$, is the surface of a circular conductor loop,d, halved. Calculate the tension, when the surface
1. is perpendicular to B
2. has an angular of $30^\circ$ with B
3. is parallel to B
How I think it could be solved
The magnetic flux
$$\Phi_B = \int_A \vec B d \vec A$$,
where $\Phi_B$ is the magnetic flux, $B$ is the magnitude of the magnetic field and $A=\pi r^2$ is the areal.

The induction tension
$$U_i = \frac{d \Phi_B}{dt} = \frac{\int_A \vec B d \vec A}{dt}$$,
where $U_i$ is the induction tension.

1. $U_i = \frac{\sin(90^\circ) A B}{dt}$
2. $U_i = \frac{\sin(30^\circ) A B}{dt}$
3. $U_i = \frac{\sin(180^\circ) A B}{dt}$
Is that correct? Could it be solved this way?

Questions
Question 1
Can I write it this way?
$$\Phi_B = \int_A \vec B d \vec A.$$
I think it looks wrong, because we have two integrals on the right side and a number on the left side. How could I write it so it doesn't look wrong? I think there are more than one way to write it correct, so please write more than one solution. I think you could use the cross product sign?

Question 2
But now I have solved it the Gaußian way(?). Could I solve it Lorentzian too? I thought of using $F=q\,\vec v \times \vec B$. The firs formular I suggested is, I think, wrong (it was about the Hall Effect).

Yours Faitfully,
and thank you very much in advance!

Last edited: Mar 7, 2015
4. Mar 7, 2015

### BvU

Dear Basip,
a quick answer I may have to correct when I'm more awake:
My compliments for your now much clearer post. You have the given variables, the right equation and the right plan to solve. So go ahead !

As to your questions: 1. yes, it is a bit strange, but correct. Check Faraday's law. That $\vec {dA}$ is a surface area .
2. I wouldn't consider this the gaussian way. That has to do with divergence. But in both cases a surface integral is needed.
There is an approach based on the Lorentz force, I think, but I can't investigate now. Check further down in the Faraday link. But it involves more math.

Last edited: Mar 9, 2015
5. Mar 9, 2015

### BvU

Everything has fallen into place ? And all done ? Or are there further questions ?

6. Mar 10, 2015

### Basip

Thank you for your help :-)