Question

A 15.0 m \long copper wire, 2.20 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter 2.10 cm (outer edge).a) What is the length of the solenoid?
b) What is the field at the center when the current in the wire is 16.7 A ?

Answer 1

a) 0.5 m

b) $4.77\cdot 10^{-3} T$

Explanation:

a)

At the beginning, the length of the copper wire is:

L = 15.0 m

and the diameter (so, the thickness of each adjacent circle) is

$t=2.20 mm = 0.0022 m$

While the diameter of one circle of the solenoid is

$d=2.10 cm = 0.021 m$

So the perimeter of one circle is

$p=\pi d=\pi (0.021)=0.0659 m$

So the number of complete circles in the solenoid is

$n=\frac{L}{p}=\frac{15.0}{0.0659}=227.6$

The tickness of one circle is $t$, so the total length of the solenoid will be:

$L'=nt=(227.6)(0.0022)=0.5 m$

b)

The magnetic field at the center of a solenoid is given by

$B=\mu_0 n I$

where

$\mu_0$ is the vacuum permeability

n is the number of turns of the solenoid

I is the current in the solenoid

Here we have:

$\mu_0 =4\pi \cdot 10^{-7}H/m$ is the vacuum permeability

$n=227.6$ is the number of turns in the solenoid (calculated in part a)

I = 16.7 A is the current in the solenoid

Substituting, we find:

$B=(4\pi \cdot 10^{-7})(227.6)(16.7)=4.77\cdot 10^{-3} T$