A solenoid is formed by winding 23.0 m of insulated silver (resistivity 1.59 * 10-8 Ω·m) wire around a hollow cylinder. The turns are wound as closely as possible without overlapping, and the insulating coat on the wire is negligibly thin. When the solenoid is connected to an ideal (no internal resistance) 3.00-V battery, the magnitude of the magnetic field inside the solenoid is found to be 6.69 * 10-3 T. Determine the radius of the wire. Hint: Because the solenoid is closely coiled, the number of turns per unit length depends upon the radius of the wire.

Given that,

Resistivity p = 1.59 * 10^-8 Ω·m

Length of coil L = 23m

Voltage across solenoid V = 3V

Magnetic field B = 6.69 * 10^-3 T.

Radius of wire r=?

Magnetic field is give as

B = μo•i•n

Where n is number of turns per length, then, n = N/L

N = L/2r

Therefore, n = 1/2r

So, B = μo•i/2r

Then, r = μo•i/2B

From ohms law

V=iR

i = V/R

So, r = μo•V/2R•B

Our resistance is given as

R = pL/A

Then, we have

r = μo•V/2R•B

r = μo•V•A/2•p•L•B

A is the area and it is given as

A = πr²

r = μo•V•πr²/2•p•L•B

Then, making r subject of formula

r = 2•p•L•B / μo•V•π

r = (2*1.59*10^-8*23*6.69*10^-3) / (4π*10^-7 * 3 * π)

r = 4.13 * 10^-4m

The radius of the wire is 4.13*10^-4m

Answer: 4.14*10^-4 m

Explanation:

n = N / L, where N = l/2r. So that

n = (l/2r) / L

n = 1/2r

Also, we know that

B = μnI. Substituting for n, we have

B = μI / 2r. If we make r subject of formula

r = μI / 2B

Recall, V = IR, and I = V/R, substituting for I

r = μV / 2BR

Also, ρ = RA/L, and R = ρL / A

substituting R also, we have

r = μVA / 2BρL

A can also be represented as, πr², so that we have

r = μVπr² / 2BρL

1 = μVπr / 2BρL, making r subject of formula again, we have

r = 2BρL / μVπ

r = (2 * 6.69*10^-3 * 1.59*10^-8 * 23) / (4π*10^-7 * 3 * π)

r = 4.89*10^-9 / 1.18*10^-5

r = 4.14*10^-4 m