Question

You have a spool of copper wire 4.82 mm in diameter and a power supply. You decide to wrap the wire tightly around a soda can that is 12.0 cm long and has a diameter of 6.50 cm, forming a coil, and then slide the wire coil off the can to form a solenoid. If the power supply can produce a maximum current of 230 A in the coil, what is the maximum magnetic field you would expect to produce in this solenoid? Assume the resistivity of copper is 1.68 ✕ 10−8 Ω · m. (Enter the magnitude.)

Answer 1

To solve the problem it is necessary to have the concepts of the magnetic field in a toroid.

A magnetic field is a vector field that describes the magnetic influence of electric charges in relative motion and magnetized materials.

By definition the magnetic field is given by the equation,

$B=\frac{\mu_0 NI}{2\pi r}$

Where,

$\mu_0$ = Permeability constant

N = Number of loops

I = Current

r = Radius

According to the given data we have that the length is 120mm and the thickness of the copper wire is 4.82mm.

In this way the number of turns N would be

$N=\frac{120mm}{4.82mm}$

$N = 24.89 \approx 25 turns$

On the other hand to find the internal radius, we know that:

$2\pi r_i = 12cm$

$r_i= \frac{12}{2\pi}$

$r_i= 1.91cm$

Therefore the total diameter of the soda would be

$r= r_i+r_o = 1.91+6.5=8.51cm$

Applying the concept related to magnetic field you have to for the internal part:

$B_i=\frac{\mu_0 NI}{2\pi r_i}$

$B_i=\frac{(4\pi*10^{-7}) (25)(230)}{2\pi (1.91*10^{-2})}$

$B_i = 0.060T$

The smallest magnetic field would be on the outside given by,

$B_o=\frac{\mu_0 NI}{2\pi r}$

$B_o=\frac{(4\pi*10^{-7}) (25)(230)}{2\pi 8.51}$

$B_o = 0.0136T$

Therefore the maximum magnetic field is 0.06T.