Area vector S

  • #1
To find the emf generated by two ten turn coils, the planes of which are at 60 degrees in a radial magnetic field B=Bcos(theta)sin(omega*t) in the direction Ar, that rotates with omega rad/sec at the instant when the coil A1A2 makes an angle alpha with the plane of the maximum flux density.

Now to compute the emf we will ignore the effect of mutual inductance and will just calculate the field for each coil then add them.

we have: emf = -N (integral) dB/dt . ds + N (integral) v x B . dL

my problem is with the first term, the S area vector is in the direction of A(phi) as shown (talking about the A1A2 coil here) so the dot product result will be zero even though the differential of B is not zero.

now i checked it with my tutor then he says we have to consider another area as the flux is clearly not zero (not that clear to me), so:
ds= r*d(phi)*dz Ar to get a value for the flux in the loop.. !!!

what i can manage so far that he chose another area that enclosed the same path, but isnt the value of that term going to increase if we choose larger areas, so the more the area "chosen" the more the flux!!.. then theres no exact magnitude for any area integral .. thanx for your help
 

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Answers and Replies

  • #2
berkeman
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Moved to homework forum.

Why is A(phi) dot B zero? In your figure, they are not at right angles to each other.
 
  • #3
the three coordinate directions am using are: A(r) A(phi) and A(z)
isnt A(r) . anything but A(r) zero?? as in A(x) . A(y) is zero

PS: i am not really asking about that particular question its the concept of intergrating over a surface that encloses a path, as explained in the previous post.

thank you
 
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  • #4
berkeman
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Lonley Sheperd said:
what i can manage so far that he chose another area that enclosed the same path, but isnt the value of that term going to increase if we choose larger areas, so the more the area "chosen" the more the flux!!.. then theres no exact magnitude for any area integral
It won't include a larger value of B dot dS. If the area is bounded by the same path, then the integral of the dot product should be the same. Like if the surface is a flat disk or a hemisphere, if the equator is the bounding path for each, then the integral of the dot product should be the same.
 
  • #5
okay.. can u give me a numerical example for that..
if B= B(t) A(r) then it will move out of the integral sign and the only term remaining is the (integral) ds , so the more the area chosen the higher the magnitude value of that term.. please help me with that
 

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