Solenoid produces a magnetic field help

  • #1
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)

 

Answers and Replies

  • #2
1,860
0
Were you looking for a correct? It's correct.
 
  • #4
1,860
0
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. :)
 
  • #5
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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