An air-filled toroidal solenoid has a mean radius of 15.2 cm and a cross-sectional area of 4.98 cm2 as shown in (Figure 1) . Picture this as the toroidal core around which the windings are wrapped to form the toroidal solenoid. The current flowing through it is 11.5 A , and it is desired that the energy stored within the solenoid be at least 0.394 J .

Question

An air-filled toroidal solenoid has a mean radius of 15.2 cm and a cross-sectional area of 4.98 cm2 as shown in (Figure 1) . Picture this as the toroidal core around which the windings are wrapped to form the toroidal solenoid. The current flowing through it is 11.5 A , and it is desired that the energy stored within the solenoid be at least 0.394 J .

Part A:
What is the least number of turns that the winding must have?

Answer: ______ turns

Answer 1

To calculate the least number of turns that the winding must have, we need to use the formula for the energy stored in an inductor (solenoid), which is given by:

E = (1/2) * L * I^2

Where:
E = Energy stored in the solenoid
L = Inductance of the solenoid
I = Current flowing through the solenoid

In this case, we are given the energy (E) as 0.394 J and the current (I) as 11.5 A. We need to find the inductance (L) in order to determine the number of turns.

However, the inductance of a toroidal solenoid depends on its geometric parameters, including the mean radius (r) and the cross-sectional area (A). The inductance can be calculated using either self-inductance or mutual inductance, depending on the configuration of the winding.

Since the problem does not provide specific information about the winding configuration, we will assume a simple solenoid where the winding is evenly distributed around the toroidal core.

In this case, we can approximate the inductance using the formula for the self-inductance of a solenoid:

L ≈ (μ₀ * N² * A) / (2 * r)

Where:
μ₀ = Permeability of free space (constant)
N = Number of turns
A = Cross-sectional area
r = Mean radius

We can rearrange the formula to solve for N:

N = √((2 * r * L) / (μ₀ * A))

Now we can substitute the given values and solve for N:

N = √((2 * 0.152 m * 0.394 J) / ((4π * 10^-7 T·m/A) * 0.0498 * 10^-4 m²))

Simplifying:

N = √(0.752 m² * T·m/A) * (0.382 Turner)

Hence, the least number of turns that the winding must have is approximately 0.382 Turner. Since turns must be a whole number, we round up to 1 turn.

Therefore, the answer is 1 turn.