Number of turns for a solenoid

  • Thread starter Matt21
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  • #1
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Homework Statement


The magnetic field in the middle of a solenoid of length L= 40cm and area of the cross-section A=12 cm^2 is B = 3 mT. The current through the solenoid is I = 4 A. What is the number of turns in this solenoid?

Homework Equations


BL=μ0*n*I
=(4πx10^-7)*n*I

The Attempt at a Solution


I have attempted this problem many times trying to use the formula above or similar variations of it with no success. I need to use the area of the cross-section in a specific formula to calculate the number of turns. Is there a formula that I'm missing as it is not in my notes, if not how would I proceed?
 

Answers and Replies

  • #2
gneill
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What do you consider to be a non successful attempt? Can you show us one? Are you sure that the solenoid cross sectional area is important in this case?

By the way, there's no RLC circuit involved here. There's no resistor (R) or capacitor (C) involved. I'll change the thread title to have it describe the actual problem.
 
  • #3
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You appear to have all the information you need. I'm not sure where your confusion is without an example of your attempt. The only thing I see is that you included L in your equation (which I assume is the length), which you shouldn't have. If you use Ampere's Law, you'll find that the magnetic field in a solenoid is only dependent on the number of turns per unit length (n) and the current running through them (I). The correct equation would therefore be ##B=\mu_0 n I##.
 
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