Question

a 2.50 mh toroidal solenoid has an average radius of 6.00 cm and a cross-sectional area of 2.00 cm2 . (a) how many coils does it have? (make the same assumption as in example 30.3.) (b) at what rate must the current through it change so that a potential difference of 2.00 v is developed across its ends?

Answer 1

The number of coils in the toroidal solenoid is calculated using its inductance, radius, and cross-sectional area. The rate of change of current is determined based on the induced potential difference.

In this scenario, we have a toroidal solenoid with an inductance L of 2.50 mH, an average radius r of 6.00 cm, and a cross-sectional area A of 2.00 cm².

(a) Calculating the Number of Coils:

The inductance of a toroidal solenoid is given by the formula:

`\[ L = (\mu_0 \cdot N^2 \cdot A)/(2\pi R) \]`

where:

- L is the inductance,

-
`\(\mu_0\)` is the permeability of free space
`(\(4\pi * 10^(-7) \, \text{T m/A}\))`,

- N is the number of coils,

- A is the cross-sectional area,

- R is the average radius.

Rearrange the formula to solve for \(N\):

`\[ N = \sqrt{(2 \pi L \cdot R)/(\mu_0 \cdot A)} \]`

Substitute the given values into the formula to calculate the number of coils.

(b) Rate of Change of Current:

The induced emf (V) in a toroidal solenoid is given by Faraday's law:

`\[ V = -N \cdot (d\Phi)/(dt) \]`

where:

- (V) is the potential difference,

- (N) is the number of coils,

-
`\((d\Phi)/(dt)\)` is the rate of change of magnetic flux.

Rearrange the formula to solve for \(\frac{d\Phi}{dt}\):

`\[ (d\Phi)/(dt) = -(V)/(N) \]`

Now, substitute the given potential difference (V) and the calculated number of coils (N) into the formula to find the rate of change of current.