Magnetic field of a solenoid problem

[Booney]

A solenoid of radius 0.0523m is made of a long piece of wire of radius 0.002432m, length 54.0m, and resistivity 7.4x10^-9 $$\Omega$$m. The solenoid is closely wound with a single wire thickness. The permeability of free space is 1.25664x10^-6 Tm/A. Find the magnetic field at the center of the solenoid if the wire is connected to a battery having an emf 26.26V. Answer in units T.

I know that B=$$\mu$$nI where $$\mu$$ is the permeability of free space, n is the number of turns per unit length (turns density), and I is the current in the wire.

I calculated the current to be:
I=V/R

V=26.26V
R=(7.4x10^-9$$\Omega$$m)/(54.0m)=1.37037x10^-10$$\Omega$$

I=1.91627x10^11 A

I have no idea how to calculate the turns density and I'm not sure if I calculated the current right either

 

[collinsmark]

Hello Booney,

Welcome to Physics Forums!

R=(7.4x10^-9$$\Omega$$m)/(54.0m)=1.37037x10^-10$$\Omega$$

Check your resistance formula. The resistance R is proportional to the resistivity, proportional to the length of the wire, and inversely proportional to the wire's cross sectional area.

[Edit: also ensure you use the length of the wire that is tightly wrapped around making the solenoid. Not the length of the solenoid itself. See below for a hint on how to find the length of the wire.]

I have no idea how to calculate the turns density

You know the solenoid's radius. So what length of wire is required for 1 turn around the solenoid? You know the wire's radius, so how many wire widths (i.e turns) does it take to span the solenoid's length?

 

[Booney]

ok so what you're saying is the resistance is [(resistivity)x(length of the wire)]/(cross-sectional area of the wire) ?

and for turns density:
the length of the wire to make one turn around the solenoid is the circumference of the solenoid.
the length of the solenoid isn't given however the length of the wire is.
so the number of turns over the length of the solenoid is (length of the wire)/(solenoid circumference) right?
then to get the length of the solenoid you use (number of turns)x(diameter of wire)?

thanks for welcoming me! I'm going to try and be active on the forum when I can since I need to brush back up on my physics, ESPECIALLY E&M (not my fav at all)

 

[collinsmark]

ok so what you're saying is the resistance is [(resistivity)x(length of the wire)]/(cross-sectional area of the wire) ?

Yes, That's right.

That being said, before you go on, could you check your resistivity number again, and make sure it is correct? A resistivity of 7.4 x 10^-9 Ω·m is half that of silver (at room temperature), and silver is the lowest resistivity you can get when using natural, non-superconducting elements. So the resistivity in the problem statement is a bit far-fetched. Are you sure the units are Ω·m rather than say, Ω·cm?

and for turns density:
the length of the wire to make one turn around the solenoid is the circumference of the solenoid.

Right!

the length of the solenoid isn't given however the length of the wire is.

Good catch. My mistake. That makes things a little easier then.

so the number of turns over the length of the solenoid is (length of the wire)/(solenoid circumference) right?
then to get the length of the solenoid you use (number of turns)x(diameter of wire)?

Yes, and that's a completely valid way to proceed. However, there is a shortcut you can take if you wanted to. Divide 1 meter by the diameter of the wire. That's gives you the number of turns per meter (for a solenoid closely wound with a single wire thickness).

thanks for welcoming me! I'm going to try and be active on the forum when I can since I need to brush back up on my physics, ESPECIALLY E&M (not my fav at all)

See you around!

 

[Booney]

awesome! thanks so much! And yeah I double checked the resistivity and the units given in the problem and I posted it correctly...This problem isn't very close to reality then huh? I didn't know that about silver though! You learn something new everyday!