Solenoid produces a magnetic field help

AI Thread Summary
A solenoid measuring 0.20 m in length with 100 turns generates a magnetic field of 1.5 mT. The current in the coil can be calculated using the formula I = BL/μN. Substituting the values, the initial calculation yields approximately 23.87 A, but this assumes an infinite solenoid. A more accurate calculation, considering finite length, gives a current of approximately 0.024 A. The discussion highlights the importance of the solenoid's assumptions in magnetic field calculations.
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1. A solenoid is 0.20 m long and consists of 100 turns of wire. At its center, the solenoid produces a magnetic field with a strength of 1.5 mT. Find the current in the coil.




2. L be the length of the solenoid = 0.2m

N be the number of turns = 100

B be the magnetic field = 1.5 mT = 1.5 x 10-3 T

I be the current in the coil in ampere

μ be the permeability in air = 4 π x 10-7 H/m



in a close circuit magnetic field



B = μNI / L




3. So, I = BL / μN

= (1.5x10-3 x 0.2) / (4 π x 10-7 x 100)

= 3 x 10-3 / 1256.637061 x 10-7

= (3/1256.637061) x 104

= 23.87324147 A

( Ans)

 
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Were you looking for a correct? It's correct.
 
Mindscrape said:
Were you looking for a correct? It's correct.

Yes, I was. And it really is? Yes!
 
Well, technically, it isn't because the derivation you used assumes an infinite solenoid. :p

It is what your professor/teacher would be looking for though. :)
 
Given that,
L = 0.20m, n = 100 turns, B = 1.5 x 10 ^-6 T
have I = B/μo . l

= 1.5 x 10^ -6 x 0.20/4 x 3.14 x 10 ^-7

=0.024 x 10 A

=2.4 10 ^-2 A
 
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