# Magnetic flow

• Feb 4th 2008, 01:10 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.
• Feb 4th 2008, 09:49 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 $\frac{dy}{dt} = - k A \sin (kt)$ .....
• Feb 5th 2008, 12:10 AM
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
• Feb 5th 2008, 12:59 AM
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.
• Feb 5th 2008, 01:10 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.

• Feb 5th 2008, 03:02 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
• Feb 5th 2008, 04:06 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.