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Chapter 30: Problem 17

An air-filled toroidal solenoid has a mean radius of 15.0 cm and a cross- sectional area of 5.00 cm\(^2\). When the current is 12.0 A, the energy stored is 0.390 J. How many turns does the winding have?

Short Answer

Expert verified
The toroidal solenoid has approximately 785 turns.

Step by step solution

01

Understand the formula for energy in a solenoid

The energy stored in an inductor (solenoid) is given by the formula:\[ E = \frac{1}{2} L I^2 \]where \( E \) is the energy, \( L \) is the inductance, and \( I \) is the current. We are given \( E = 0.390 \) J and \( I = 12.0 \) A.
02

Understand the formula for inductance of a toroidal solenoid

The inductance \( L \) of a toroidal solenoid is calculated using the formula:\[ L = \mu_0 \frac{N^2 A}{2 \pi r} \]where \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \) is the permeability of free space, \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( r \) is the mean radius.
03

Substitute known values to relate energy with inductance

We have the energy formula \( 0.390 = \frac{1}{2} L (12.0)^2 \) and substituting for \( L \) from the inductance formula:\[ 0.390 = \frac{1}{2} \left( \mu_0 \frac{N^2 A}{2 \pi r} \right) (12.0)^2 \].
04

Rearrange formula to solve for number of turns \( N \)

Isolate \( N^2 \) from the equation:\[ L = \frac{2E}{I^2} = \mu_0 \frac{N^2 A}{2 \pi r} \]Rewriting it:\[ N^2 = \frac{2E \cdot 2 \pi r}{\mu_0 \cdot A \cdot I^2} \]Substitute values: \( E = 0.390 \), \( A = 5.00 \times 10^{-4} \) m\(^2\), \( r = 0.150 \) m, \( I = 12.0 \).
05

Calculate \( N \) from rearranged equation

Substitute in the constant values and do the math:\[N^2 = \frac{2 \times 0.390 \times 2 \pi \times 0.150}{4\pi \times 10^{-7} \times 5.00 \times 10^{-4} \times (12.0)^2}\]Simplifying this gives \( N^2 = 617500 \), thus \( N \approx 785 \) turns.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductance
Inductance is a property of an electrical conductor which opposes a change in the current flowing through it. In simple terms, it tells us how much electromagnetic energy can be stored in the conductor when a current passes through it. Inductance is usually denoted by the symbol \(L\). For a toroidal solenoid, the formula to determine inductance is \( L = \mu_0 \frac{N^2 A}{2 \pi r} \). Here:
  • \( \mu_0 \) is the permeability of free space, a constant value of \( 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \).
  • \(N\) is the number of turns which impacts how much inductance the solenoid can have.
  • \(A\) refers to the cross-sectional area of the solenoid, impacting the electromagnetic field produced.
  • \(r\) is the mean radius of the toroid which also affects the field strength.
All these components combined give us an idea of how efficiently the solenoid can store magnetic energy.
Energy Stored in Solenoid
The energy stored in a solenoid is a measure of how much electromagnetic energy can be accumulated due to the inductance and current flowing through it. The formula \(E = \frac{1}{2} LI^2\) is used for this calculation. Here:
  • \(E\) is the energy stored in Joules.
  • \(L\) is the inductance, which we've discussed above.
  • \(I\) is the current flowing through the solenoid, measured in Amperes.
This relationship shows that the energy stored in the solenoid is proportional to both the inductance and the square of the current. As the current increases, the energy stored increases significantly; thus, making understanding this formula crucial for applications involving high currents.
Magnetic Fields
Magnetic fields in solenoids are created when an electric current passes through the coil. In a toroidal solenoid, the magnetic field is confined within the core, creating a uniform field along the coil's length. Here are some important points:
  • The direction of the magnetic field is determined by the current direction, following the right-hand rule.
  • The strength of this field is directly proportional to the number of turns \(N\), the current \(I\), and the permeability \(\mu_0\).
  • The field strength becomes stronger with more tightly packed coils or a higher current.
This magnetic characteristic is why solenoids are employed in devices that need precise control of magnetic fields, such as inductors, electromagnets, and transformers.
Number of Turns
The number of turns \(N\) in a toroidal solenoid significantly influences its inductance and the magnetic field it produces. More turns generally mean greater inductance and stronger magnetic fields. This is because each loop of the coil contributes to the overall magnetic field. In the exercise, the number of turns \(N\) was found using the relationship between the known energy stored, current, and inductance:
  • Using the equation \(N^2 = \frac{2E \cdot 2 \pi r}{\mu_0 \cdot A \cdot I^2}\).
  • Substituting the given values, we determined \(N\) to be approximately 785 turns.
Thus, when designing solenoids for specific applications, calculating the appropriate number of turns is crucial to achieving the desired magnetic properties.

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Most popular questions from this chapter

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_1\) turns. A second toroidal solenoid with \(N_2\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_1 = 500\) turns, \(N_2 = 300\) turns, \(r = 10.0\) cm, and \(A = 0.800 \, \mathrm{cm}^2\), what is the value of the mutual inductance?

A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d\Phi_B\) through a narrow strip of length \(l\) parallel to the axis, of width \(dr\), at a distance \(r\) from the axis of the cable and lying in a plane containing the axis. (c) Integrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(l\) of the cable is $$L = l \frac{\mu_0}{2\pi} \mathrm{ln}(\frac{b}{a})$$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length l of the cable.

A solenoid 25.0 cm long and with a cross-sectional area of 0.500 cm\(^2\) contains 400 turns of wire and carries a current of 80.0 A. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

An inductor with an inductance of 2.50 H and a resistance of 8.00 \(\Omega\) is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current.

The minimum capacitance of a variable capacitor in a radio is 4.18 pF. (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\)-\(C\) circuit is \(1600 \times 10^3\) Hz, corresponding to one end of the \(\textbf{AM}\) radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^3\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?
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