What Is the Magnetic Field at the Center of a Solenoid?

Click For Summary
SUMMARY

The magnetic field at the center of a solenoid can be calculated using the formula B = μ(N/L)I, where μ is the permeability of free space, N is the number of turns, L is the length of the solenoid, and I is the current. In this discussion, a solenoid with a length of 0.25m, radius of 0.025m, and 440 turns carrying a current of 12 A yields a magnetic field of 0.0265 T. However, the teacher's expected answer of 2.21 x 10-3 T suggests a potential misunderstanding regarding whether the calculation should focus on magnetic flux density or field strength per amp of current.

PREREQUISITES
  • Understanding of solenoid physics
  • Familiarity with the formula B = μ(N/L)I
  • Knowledge of magnetic field concepts
  • Basic algebra for calculations
NEXT STEPS
  • Research the concept of magnetic flux density versus magnetic field strength
  • Learn about the permeability of free space and its role in magnetic calculations
  • Explore variations in solenoid design and their effects on magnetic fields
  • Investigate common misconceptions in electromagnetism education
USEFUL FOR

Students studying electromagnetism, physics educators clarifying solenoid concepts, and anyone interested in understanding magnetic field calculations in practical applications.

kiltfish
Messages
6
Reaction score
0
This is driving me insane. This is all I can think of to do, and my teacher said it was wrong.

A solenoid of length 0.25m and radius 0.025m is comprised of 440 turns of wire. Determine the magnitude of the magnetic field at the center of the solenoid when it carries a current of 12 A.



B=u(N/L)I

B=0.0265 T
 
Physics news on Phys.org
I don't see what's wrong with that. Unless your teacher wants you to compute the magnetic flux density rather than the field strength?
 
Apparently he insists that the answer is 2.21 x 10-3 T. I don't think that's what you would get calculating for the magnetic flux density, but if it is, how can you tell he's asking for that is the problem? I gave you the exact words he asked.
 
Let's see how his answer compares to the one you (and I) calculate:

(2.654 x 10^-2 T)/(2.21 x 10^-3 T) = 12.0

His answer seems to be 1/12 of the value that we think it should be. Could it be that he wanted the field strength per amp of current?
 

Similar threads

Calculating magnetic field outside a solenoid
  • · Replies 15 ·
Replies
15
Views
1K
Non-Uniform Magnetic Field Calculations
  • · Replies 9 ·
Replies
9
Views
2K
Why are the effects of a magnetized material neglected in this case?
  • · Replies 5 ·
Replies
5
Views
2K
Why is magnetic field half at the sides of a solenoid?
  • · Replies 11 ·
Replies
11
Views
3K
Understanding self-inductance in a solenoid.
  • · Replies 3 ·
Replies
3
Views
2K
Electron in a time variable magnetic field
  • · Replies 5 ·
Replies
5
Views
2K
{ help } Calculating the B-field at the center of a solenoid
  • · Replies 1 ·
Replies
1
Views
2K
Current in circular wire surrounding infinite solenoid in a circuit
  • · Replies 7 ·
Replies
7
Views
1K
Direction and magnitude of electric field produced by current change
  • · Replies 8 ·
Replies
8
Views
1K
Solenoid with current carrying wire inside
  • · Replies 3 ·
Replies
3
Views
3K