I KEPT GETTING THE WRONG ANSWER??? DID I DO ANYTHING WRING? A 32.0−cm long solenoid 1.25 cm in diameter is to?
A 32.0−cm long solenoid 1.25 cm in diameter is to produce a field of 0.275T at its center.
How much current should the solenoid carry if it has 900 turns of the wire?
I used the formula I = B*D/(µ0*N)
So it would be (0.275*0.0125)/(4π*10^-7*900) = 3.039 A
But that's not the answer.
Did I do anything wrong on this problem?
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I quickly looked up the formula online and it looks as though D should read "L" in your formula.
Here are some of the sites I found:
http://hyperphysics.phy-astr.gsu.edu/hbase/magneti...
http://hyperphysics.phy-astr.gsu.edu/hbase/magneti...
http://www.calctool.org/CALC/phys/electromagnetism...
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/...
If the center of solenoid is air, then it is u0 rather than k(u0) in formula
I= B/(u0)(n). .... n is the turn density...
n=N/L (number of turns)/(length of coil)
I= (B)(L) / (u0)(N)
Where N is number of turns, L is the length of coil in meters
I =(0.275)(0.32)/(4pi*10^-7)(900)
=77.8A
I just verified the answer with the calculator in the third link.
Also, the sites are saying that the above formula is accurate for coil lengths that are significantly larger than the radius of coil. The site with the calculator states that the radius is not required, and even when I change the radius in the calculator, it does not change the calculation of B, which means that it would not change the calculation of Current. So this should make you feel better about not using the radius in your calculation.
Hope that helps!