Problem 47 - University Physics with Modern Physics

Question

A toroidal solenoid with 500 turns is wound on a ring with a mean radins of 290 \(\mathrm{cm}\) . Find the current in the winding that is required to set up a magnetic field of 0.350 \(\mathrm{T}\) in the ring (a) if the ring is made of annealed iron \(\left(K_{m}=1400\right)\) and \((b)\) if the ring is made of sillicon steel \(\left(K_{m}=5200\right)\) .

Step-by-Step Solution

Answer

(a) 7.21 A for annealed iron, (b) 1.94 A for silicon steel.

Step 1: Understanding the Relationship

The magnetic field inside a toroidal solenoid is given by the formula \( B = \mu \cdot n \cdot I \), where \( B \) is the magnetic field, \( \mu \) the permeability of the material, \( n \) the number of turns per unit length, and \( I \) the current. We also know that \( \mu = \mu_0 \cdot K_m \), where \( \mu_0 = 4\pi \times 10^{-7} \ \text{T\cdot m/A} \) is the permeability of free space and \( K_m \) is the relative permeability.

Step 2: Calculate the Number of Turns Per Unit Length (n)

The mean circumference of the ring (the path of the coil) is \( 2\pi r \). With a mean radius \( r = 290 \ \text{cm} = 2.9 \ \text{m} \), the mean circumference is \( 2\pi \times 2.9 \ \text{m} \approx 18.22 \ \text{m} \). The number of turns per unit length, \( n \), is therefore \( n = \dfrac{500}{18.22} \approx 27.45 \ \text{turns/m} \).

Step 3: Calculate the Current for Annealed Iron

For annealed iron, \( K_m = 1400 \). Therefore, \( \mu = \mu_0 \cdot 1400 = 4\pi \times 10^{-7} \cdot 1400 = 1.76 \times 10^{-3} \ \text{T\cdot m/A} \). Substituting into the magnetic field equation, we have \( 0.350 = 1.76 \times 10^{-3} \cdot 27.45 \cdot I \). Solving for \( I \), \( I = \dfrac{0.350}{1.76 \times 10^{-3} \times 27.45} \approx 7.21 \ \text{A} \).

Step 4: Calculate the Current for Silicon Steel

For silicon steel, \( K_m = 5200 \). Therefore, \( \mu = \mu_0 \cdot 5200 = 4\pi \times 10^{-7} \cdot 5200 = 6.53 \times 10^{-3} \ \text{T\cdot m/A} \). Using the same approach, \( 0.350 = 6.53 \times 10^{-3} \cdot 27.45 \cdot I \). Solving for \( I \), \( I = \dfrac{0.350}{6.53 \times 10^{-3} \times 27.45} \approx 1.94 \ \text{A} \).

Key Concepts

Magnetic Field

The magnetic field is a crucial concept when dealing with a toroidal solenoid, a device widely used in electronic components like transformers and inductors. In a toroidal solenoid, the magnetic field is contained within its core, producing no external magnetic field and minimizing electromagnetic interference.
The strength of this magnetic field, denoted by \( B \), is influenced by several factors:

• The current \( I \) flowing through the coil.
• The permeability \( \mu \) of the material surrounding the coil.
• The number of turns per unit length \( n \) of the solenoid.

Typically, the formula used to calculate the magnetic field inside a toroidal solenoid is \( B = \mu \cdot n \cdot I \). This equation highlights how the magnetic field is directly proportional to the current and the number of windings (or turns) in the coil, as well as the permeability of the material making up the core.

Permeability

Permeability \( \mu \) is the measure of how easily a material can support the formation of a magnetic field within itself. It is a critical parameter in the study of toroidal solenoids. The standard expression for permeability is \( \mu = \mu_0 \cdot K_m \), where \( \mu_0 \) is the permeability of free space, valued at \( 4\pi \times 10^{-7} \, \text{T\cdot m/A} \), and \( K_m \) is the dimensionless relative permeability of the material.
The relative permeability \( K_m \) compares the material's magnetic permeability to that of a vacuum. Materials with high \( K_m \), like annealed iron and silicon steel, allow for a stronger magnetic field with a smaller amount of current, making them effective in the design of magnetic circuits.

• Annealed iron: \( K_m = 1400 \)
• Silicon steel: \( K_m = 5200 \)

These values indicate how significantly different materials will impact the needed current to achieve a desired magnetic field strength, showcasing the critical role permeability plays in circuit design.

Turns per Unit Length

The concept of turns per unit length \( n \) refers to how densely the coil is wound along its path. It is expressed as the number of turns (or loops) of wire per meter along the solenoid. This density of winding is essential in determining the magnetic field strength within the solenoid.
In our particular exercise involving a toroidal solenoid, the number of turns per unit length was calculated based on the mean circumference of the coil's path: \( n = \frac{500}{18.22} \approx 27.45 \ \text{turns/m} \).
The more densely the coil is wound, the stronger the magnetic field inside the solenoid, given the same current and material permeability. This concept underlines the importance of designing the solenoid with appropriate winding density to meet specific magnetic field requirements without unnecessarily increasing current.