Question

An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 4.95 cm2 . When the current is 12.5 A , the energy stored is 0.390 J . You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Storing energy in an inductor. Part AHow many turns does the winding have?

Answer 1

I will try to define the net problem, without the intermediate message between the message as:

An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 4.95 cm2 . When the current is 12.5 A , the energy stored is 0.390 J .

Part A: How many turns does the winding have?

To solve the problem it is necessary to apply the concepts related to the storage of energy in an inductor and how it is possible to calculate from the inductance the number of turns of the system.

By definition we know that the energy stored in an inductor is given by,

$E = \frac{1}{2} LI^2$

Where,

L = Inductance

I = Current

In this way, clearing the Inductance in the previously given equation we have to

$L = \frac{2E}{I}$

$L = \frac{2(0.39J)}{12.5A}$

$L = 0.0624H$

In a system the inductance is given by

$L =\frac{\mu_0 N^2A}{l}$

Where l represents the length, however as we deal with the perimeter of a circle we have,

$L =\frac{\mu_0 N^2A}{2\pi R}$

Replacing our values we have

$(0.0624)=\frac{(4\pi*10^{-7})(N)^2(12.5)}{2\pi (15.5/2*10^{-3})^2}$

Re-arrange to find N,

$N^2 = \frac{(0.0624)2\pi (15.5/2*10^{-3})}{(4\pi*10^{-7})(4.95*10^{-4})}$

$N^2 = 4884848.48$

$N = 2210.16\approx 2211 turns$

Therefore the winding have 2211turns