Magnetic flow

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February 4th 2008, 12:10 PM
JBswe

Magnetic flow

Hiya need a little help with this please

The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

I think i need to derv M=B*Acos(100pi*T) and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

if anyone could give me a hand i would be thankful.
February 4th 2008, 08:49 PM
mr fantastic

Quote:

Originally Posted by JBswe

Hiya need a little help with this please

The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

I think i need to derv M=B*Acos(100pi*T) Mr F says: Don't think, know. You have obviously studied induced E.M.F. and Faraday's Law ......

and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

if anyone could give me a hand i would be thankful.

If $y = A \cos (kt)$ then $\frac{dy}{dt} = - k A \sin (kt)$ .....
February 4th 2008, 11:10 PM
JBswe

Ok so M=B*Acos(100pi*T)

dy/dt = -100*Asin(100T)

A is = 2.5
B is =150

I swap in the values

= -100*150*2.5sin(100T)

so the amplitude of the AC is 37500? That seems wrong
February 4th 2008, 11:59 PM
mr fantastic

Quote:

Originally Posted by JBswe

Hiya need a little help with this please

The Magnetic flow; M, through a rektangular coil with the area A kan be discribled with M=B*Acos(100pi*T) where B is the magnetic flows density. when the coil rotates - it has 50 rotations per second - inducing a AC current that gives dm/dt. find the amplitude of the AC if its magnetic flows density B=150 mT and the area A=2.5 dm^2

I think i need to derv M=B*Acos(100pi*T) and then put in the numbers but am unsure how do derv this. maybe somthing like M=B*-Asin(100pi*T)+100 ?

if anyone could give me a hand i would be thankful.

$\frac{dM}{dt} = -100 \pi B A \sin(100 \pi T)$ so the amplitude is $100 \pi B A$ .

But you've gotta get the units right, sport:

B = 150 mT = $150 \times 10^{-3}$ = 0.150 Tesla.

A = 2.5 dm^2 = $2.5 \times 10^{-2}$ = 0.025 m^2.

Note: 1 dm = 1/10 m therefore 1 dm^2 = 1/100 m^2.

So the amplitude is $(100)(\pi)(0.150)(0.025) = (15)(\pi)(0.025) = ......$
And the unit will be ....?

Wrong units mean that answers can be wrong by many orders of magnitude. Make it your business to get them right.
February 5th 2008, 12:10 AM
mr fantastic

By now you know how to get the correct answer. Just so you understand your mistakes:

Quote:

Originally Posted by JBswe

Ok so M=B*Acos(100pi*T)

dy/dt = -100*Asin(100T) Mr F says: There's a $\pi$ missing from the amplitude ... The 'k' in $M = B A \cos( 100 \pi T)$ is $100 \pi$ .

A is = 2.5 Mr F says: What are the units!!

B is =150 Mr F says: What are the units!!

The unit problem I pointed out earlier means that your answer will be out by a factor of $10^{-5}$ . But then things have to get multiplied by the missing $\pi$ .

I swap in the values

= -100*150*2.5sin(100T)

so the amplitude of the AC is 37500? That seems wrong Mr F says: At least you realised .... many students would not have.
February 5th 2008, 02:02 AM
JBswe

ok units have always been a problem for me but i think the answer should be in mT
$ \frac{dM}{dt} = -100 \pi B A \sin(100 \pi T) $

$ \frac{dM}{dt} = -100 \pi 150*10^{-3}*2.5*10^{-2} \sin(100 \pi T) $

so the amplitude is 1.18 mT? but that seems very small. maybe 1.18 T
February 5th 2008, 03:06 AM
mr fantastic

Quote:

Originally Posted by JBswe

ok units have always been a problem for me but i think the answer should be in mT
$ \frac{dM}{dt} = -100 \pi B A \sin(100 \pi T) $

$ \frac{dM}{dt} = -100 \pi 150*10^{-3}*2.5*10^{-2} \sin(100 \pi T) $

so the amplitude is 1.18 mT? but that seems very small. maybe 1.18 T

Correct value but wrong unit.

All data has been put into the equation in SI units (m^2, T). So the answer will be in SI units. And the SI unit of E.M.F (induced or otherwise) is ..... drum roll please ....... VOLT.

So the amplitude is 1.18 Volt.