Solenoid homework question

[ahazen]

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

 

[collinsmark]

Hello ahazen,

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

 

[ahazen]

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).

 

[ahazen]

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

 

[rwisz]

and u0=4*pi*10^-7, but when I solve for N, I get a very large number (approximately 3.5 x 10^10).

Firstly, it is important to note that the formula B=n*u0*I is for the magnetic field strength at the center of a solenoid. In order to calculate the number of turns of wire on a solenoid, we need to use the formula N=n*I*L, where N is the number of turns, n is the number of turns per unit length, I is the current, and L is the length of the solenoid.

Using this formula, we can rearrange it to solve for n, which is the number of turns per unit length. Substituting the given values, we get n= B/(u0*I)= (2.9 × 10-3 T)/(4*pi*10^-7 * 4.9 A) = 1.48 x 10^4 turns/m.

Next, we need to find the total number of turns on the solenoid by multiplying n by the length of the solenoid. This gives us N= (1.48 x 10^4 turns/m) * (248 cm/100 cm) = 3.67 x 10^6 turns.

Therefore, the solenoid would have approximately 3.67 million turns of wire in order to carry a current of 4.9 A and have a magnetic field strength of 2.9 × 10-3 T at its center. I would recommend double checking your calculations to ensure accuracy.