An air-filled toroidal solenoid has a mean radius of 15.0 cm and a cross-sectional area of 5.00 cm2. When the current is 12.0 A, the energy stored is 0.390 J. How many turns does the winding have?

Ch 30: Inductance

Young & Freedman Calc - University Physics 15th Edition

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Young & Freedman Calc 15th Edition

Ch 30: Inductance

Problem 15

Chapter 30, Problem 15

An air-filled toroidal solenoid has a mean radius of 15.0 cm and a cross-sectional area of 5.00 cm². When the current is 12.0 A, the energy stored is 0.390 J. How many turns does the winding have?

Verified step by step guidance

Start by recalling the formula for the energy stored in the magnetic field of a solenoid: \( U = \frac{1}{2} L I^2 \), where \( U \) is the energy stored, \( L \) is the inductance, and \( I \) is the current.

Rearrange the formula to solve for the inductance \( L \): \( L = \frac{2U}{I^2} \). Substitute \( U = 0.390 \ \text{J} \) and \( I = 12.0 \ \text{A} \) into the equation to calculate \( L \).

Next, use the formula for the inductance of a toroidal solenoid: \( L = \mu_0 \frac{N^2 A}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \ \text{T·m/A} \)), \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( r \) is the mean radius.

Rearrange the formula to solve for \( N \): \( N = \sqrt{\frac{L \cdot 2 \pi r}{\mu_0 A}} \). Substitute \( L \) from step 2, \( r = 15.0 \ \text{cm} = 0.150 \ \text{m} \), and \( A = 5.00 \ \text{cm}^2 = 5.00 \times 10^{-4} \ \text{m}^2 \) into the equation.

Simplify the expression to calculate \( N \), the number of turns in the winding. Ensure all units are consistent (meters, henries, etc.) during the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Toroidal Solenoid

A toroidal solenoid is a coil of wire shaped like a doughnut, where the magnetic field is confined within the loop. The magnetic field strength inside a toroidal solenoid depends on the number of turns per unit length and the current flowing through the wire. This configuration allows for a uniform magnetic field and is often used in applications requiring compact magnetic fields.

Magnetic Energy Storage

The energy stored in a magnetic field can be calculated using the formula U = (1/2)LI², where U is the energy, L is the inductance, and I is the current. In a toroidal solenoid, the inductance depends on the geometry of the solenoid, including the number of turns, the mean radius, and the cross-sectional area. Understanding this relationship is crucial for determining how much energy is stored based on the current and the solenoid's physical characteristics.

Inductance Calculation

Inductance is a measure of how much magnetic flux is generated for a given current in a coil. For a toroidal solenoid, the inductance can be calculated using the formula L = (μ₀N²A)/2πr, where μ₀ is the permeability of free space, N is the number of turns, A is the cross-sectional area, and r is the mean radius. This formula is essential for solving problems related to the number of turns in a solenoid when other parameters are known.

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