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

To find the number of turns in the winding of a toroidal solenoid, use the energy stored in the magnetic field and mechanical properties of the toroid to first solve for inductance and then use that to solve for the number of turns.

Step-by-step explanation:

The question asks us to calculate the number of turns in the winding of an air-filled toroidal solenoid given the energy stored, the current, the cross-sectional area, and the mean radius of the toroid. We can use the known values to rearrange the energy stored in an inductor equation, Eind = (1/2) · L · I^2, where Eind is the energy stored, L is the inductance, and I is the current. To find the inductance (L), we can use the formula for the inductance of a toroid which involves the number of turns N, the permeability of free space (µ_0 = 4 x 10^-7 Tm/A), the cross-sectional area A, and the mean radius r.

By substituting the values into the rearranged energy stored formula, we can solve for L and infer the number of turns N that the winding has using the inductance of a toroid formula.

Answer 2

Answer: 2800 turns

Step-by-step explanation:

We are given-

r = 0.155m (radius, converted from cm to m)

A = 0.000495m^2 (cross-sectional area, converted from cm^2 to m^2)

I = 12.5 A (current)

U = 0.390 J (energy stored)

We need to find-

N, number of turns.

Equations-

To get N we can use the equation for self inductance of a toroid:

`L =(u_(0)N^2A )/(2\pi r)` rearranging this for N we get:
`N=\sqrt{ (2\pi rL)/(u_(0)A ) }`

We have been given everything except L here, so we can solve for that using the stored energy formula:

`U=(LI^2)/(2)` rearrange this for L:
`L = (2U)/(I^2)`

Now, plug and solve:

`L =(2*0.390)/(12.5^2) = 0.004992`

`N=\sqrt{(2\pi(0.155)(0.004992))/((4\pi*10^(-7))(0.000495)) }=2795.67`

Therefore, the number of turns is 2800.