How Many Turns of Wire Are Needed for a Solenoid with These Specifications?

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To determine the number of wire turns needed for a solenoid with a diameter of 17 cm, a length of 248 cm, and a current of 4.9 A producing a magnetic field of 2.9 × 10^-3 T, the formula B = n * μ0 * I is used. The variable n represents the number of turns per unit length, calculated by dividing the total number of turns N by the solenoid's length. After some calculations and adjustments, the correct number of turns was found to be 1168. The initial confusion stemmed from varying interpretations of the length measurement. Ultimately, the correct approach led to the accurate determination of the solenoid's specifications.
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How many turns of wire would be on a solenoid carrying a current 4.9 A if the solenoid is 17 cm in diameter, 248 cm long, and the field at the center is 2.9 × 10-3 T?

I can't seem to get the right answer...

I plug the values into B=n*u0*I and then solve for N.
I know that n=N/length
 
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Hello ahazen,

What answer are you getting? (You need to show your work. :wink:)
 


I plug in: 2.9e-3=n(4*pi*10^-7)(4.9)
and then solve for n.
Then i divide by the length.

I keep getting something like 1899, 1.899, 18.99 depending on how i write the length (248, or .248, or .0248).
 


Thank you for your help:) I figured it out.:) I took B*L=N u0*I and solve for N to get 1168 turns.
 
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