Emf of a Closed Loop of Wire In a Magnetic Field

[rocketgirl93]

]

Homework Statement

A closed loop of wire 7.2 x10$^{}-3$ m$^{}2$ is placed so that it is at an angle of 60degrees to a uniform magnetic field. The flux density is changing at 0.1 T/s.
The emf, in V, induced in the loop of wire is
A) 3.6x10$^{}-4$
b)3.6x10$^{}-2$
6.9

Homework Equations

emf = NBA/t F=BiL

The Attempt at a Solution

Area x 0.1 x sin60

 

[I like Serena]

Welcome to PF, rocketgirl!

]

Homework Statement

A closed loop of wire 7.2 x10$^{}-3$ m$^{}2$ is placed so that it is at an angle of 60degrees to a uniform magnetic field. The flux density is changing at 0.1 T/s.
The emf, in V, induced in the loop of wire is
A) 3.6x10$^{}-4$
b)3.6x10$^{}-2$
6.9

Homework Equations

emf = NBA/t F=BiL

The Attempt at a Solution

Area x 0.1 x sin60

What do you get if you calculate Area x 0.1 x sin60?


Btw, your formula emf = NBA/t is not quite correct.
It should be:
$$emf = -N {d(BA) \over dt}$$
where BA is the product of B and the projection of the area that is perpendicular to B.

 

[rocketgirl93]

Thanks!

I know I didn't write it out properly don't worry, I was just in a hurry to get it down and figured everyone would know what I meant.

I got 6.2 x10$^{}-4$

I have the answer, and I worked backwards from that. The method seems to be
7.2 x10$^{}-3$ m$^{}2$ x 0.1
then divide this number by 2 to get the answer, 3.6 x 10$^{}-4$

I can see where the first step comes from but I don't understand why it is divided by two. I think it has something to do with because it's at sin60 not all of it perpendicular to the field. I'm not sure!

 

[I like Serena]

Yes, it has to do with sin60.

Your problem states "at an angle of 60degrees to a uniform magnetic field".

How do you know it has to be the sine and not the cosine?
Or perhaps you should use the sine of 30?

 

[rocketgirl93]

Oh! You're right! I assumed it was sin60 because of F=BILsinx
Can you please explain to me how it works with sin30 instead?
Thank you! :)

 

[I like Serena]

Well, let's draw a picture:

The closed loop is the fat line.
It is placed so that it is at an angle of 60degrees to a uniform magnetic field.

Now what is the fraction of the fat line that is perpendicular to the magnetic field?

 

[rocketgirl93]

I'm sorry, I don't know

 

[rocketgirl93]

Would you use trig to find the perpendicular component?

 

[I like Serena]

Would you use trig to find the perpendicular component?

Yes.

Do you know how the sine of an angle is defined?

 

[rocketgirl93]

sinx = O/H

But then do you have to find the diameter of the loop to act as the hypotenuse?

 

[I like Serena]

All correct. :)

Let's assume for now that you know the diameter of the closed loop.
So if you look at the triangle in the drawing, can you find the length of the short edge of the triangle?

 

[rocketgirl93]

Yes, would you do diameter x cos60?

 

[I like Serena]

Exactly!

Note that cos60=sin30=1/2.

 

[rocketgirl93]

Great! Thank you!

 

[rocketgirl93]

I tried doing it again but so far all my triangles still tell me to do sin60... and using the diameter of the circle didnt work

 

[I like Serena]

If you look at the big triangle in the drawing, its sharpest angle is 30 and the other angle is 60.

The sine of the sharp angle (30 degrees) is the short edge divided by the diameter of the loop...

Which triangles are you looking at?

 

[rocketgirl93]

I tried to draw my own

 

[I like Serena]

Aha!
You selected another angle to be 60 degrees.

I have to admit that it is a bit ambiguous which angle should be 60 degrees exactly.

In my interpretation they started with the closed loop perpendicular to the magnetic field.
And then they turned the loop 60 degrees.
That way you get the picture I drew.In your picture you apparently started with the loop parallel to the field, and then started turning.
I'm guessing that this is not what they intended.

 

[rocketgirl93]

Ohhh ok and I started parallel to the field ohh ok. Gosh this question is driving me mad. Thank you for all your help though!

 

[I like Serena]

All good now?

(Nice picture btw.)

 

[rocketgirl93]

Well all good in the sense that I have tried as hard as I possibly can with this question, and am now just accepting that my answer doesn't match up with the one it supposedly is! thank you :)