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Chapter 30: Problem 16

An air-filled toroidal solcnoid has 300 turns of wire, a mean radius of \(12.0 \mathrm{~cm}\), and a cross-sectional area of \(4.00 \mathrm{~cm}^{2}\). If the current is \(5.00 \mathrm{~A}\), calculate: (a) the magnetic ficld in the solenoid; (b) the selfinductance of the solcnoid: (c) the cnergy stored in the magnctic ficld: (d) the cnergy density in the magnetic ficld. (c) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

Short Answer

Expert verified
After following all steps, we will get the magnetic field, the self inductance, the energy stored in the magnetic field, and the energy density of a toroidal solenoid. Also, we will confirm the calculated energy density with our previous calculation.

Step by step solution

01

Calculating the Magnetic Field

The magnetic field in a solenoid is given by \(B = \mu n I\). Here, \(\mu=4 \pi \times 10^{-7}\) T m/A is the vacuum permeability, n is the number of turns per unit length, and I is the current. To calculate n, we need to take the ratio of the total number of turns to the total length: \[n=N/l\]. As we know, the total length is same as the circumference of the toroid which is given by \(2 \pi r\). After substituting the given values in the formula it gives us the magnetic field.
02

Calculating the Self Inductance

The self inductance of a solenoid is given by \(L = \mu n^2 A l\). After substituting the calculated 'n' value, the given cross-sectional area 'A' and the length 'l', and \(\mu\) as the vacuum permeability, we can calculate the self inductance.
03

Determining the Energy Stored in the Magnetic Field

The energy stored in a magnetic field is given by \(U = \frac{1}{2} L I^2\). Using the calculated 'L' as the self inductance and the given 'I' as the current, we will calculate the energy stored in the magnetic field
04

Calculating the Energy Density

The energy density is given by the ratio of the energy stored to the volume of the solenoid. The volume of a toroidal solenoid can be calculated by multiplying its cross-sectional area 'A' with its total length 'l'. We will then divide the energy stored by the volume to get the energy density. Finally, we will confirm this result by dividing the calculated stored energy by the calculated volume of the solenoid.

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Key Concepts

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

Self Inductance
Understanding self-inductance in a toroidal solenoid involves recognizing how it resists the change in current within the solenoid itself. It is represented by the symbol 'L' and can be thought of as an electrical property that influences the magnetic field created by the solenoid.
In a toroidal solenoid, the self-inductance is calculated using the formula:
  • \( L = \mu n^2 A l \)
Here:
  • \( \mu \) is the permeability of free space (\(4 \pi \times 10^{-7} \) T m/A)
  • \( n \) is the number of turns per unit length
  • \( A \) is the cross-sectional area
  • \( l \) is the length or circumference of the toroid
This formula illustrates that self-inductance depends on the geometry of the solenoid and the number of turns. Toroidal solenoids are particularly efficient due to the closed-loop shape, which minimizes energy losses.
Energy Stored in Magnetic Field
The energy stored in the magnetic field of a solenoid is an important concept, especially in understanding how magnetic fields store energy in electromagnetic devices. For a toroidal solenoid, the energy stored, denoted by \( U \), can be determined using the formula:
  • \( U = \frac{1}{2} L I^2 \)
Where:
  • \( L \) is the inductance of the solenoid
  • \( I \) is the current passing through the solenoid

This equation shows that the energy stored is proportional to both the inductance and the square of the current. It highlights how energy increases significantly as current rises. In practical applications, this stored energy is essential for the operation of inductive components like transformers and inductors.
Energy Density
Energy density in a toroidal solenoid gives insights into how much energy is stored per unit volume of the space occupied by the magnetic field. It's calculated by dividing the energy stored by the volume of the solenoid. This is given by:
  • Energy Density = \( \frac{U}{V} \)
Where:
  • \( U \) is the stored energy
  • \( V \) is the volume of the solenoid, calculated as \( A \times l \)

This concept is particularly useful when evaluating the efficiency of magnetic storage in various designs. Understanding energy density helps in enhancing the storage capabilities of electromagnetic devices, ensuring they perform optimally with minimal space requirements. By confirming that this ratio matches the calculated value, one ensures that energy sharing and distribution are accurately perceived.

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Most popular questions from this chapter

An Wectromagnetic Car Alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of \(3500 \mathrm{~Hz}\). To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be \(12.0 \mathrm{~V}\). To produce a sufticiently loud sound, the capacitor must store \(0.0160 \mathrm{~J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

If part of the magnet develops resistance and liquid helium boils away, rendering more and more of the magnct nonsuperconducting. how will this quench affect the time for the current to drop to half of its initial value? (a) The time will be shorter bccause the resistance will increase; (b) the time will be longer because the resistance will increase: (c) the time will be the same; (d) not enough information is given.

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as 0.4 T. (By comparison, the earth's magnetic field is about \(1 / 10,000\) as strong. ) Sunspots can be as large as \(25,000 \mathrm{~km}\) in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\). Assume \(\mu\) for the sunspot material is \(\mu_{0}\). If \(100 \%\) of the magneticfield energy stored in a sunspot could be used to eject the sunspot's material away from the sun's surface, at what speed would that material be cjected? Compare to the sun's escape speed, which is about \(6 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) (Hint: Calculate the kinetic energy the magnetic field could supply to \(1 \mathrm{~m}^{3}\) of sunspot material.)

I-C Oscillations. A capacitor with capacitance \(6.00 \times 10^{-5} \mathrm{~F}\) is charged by connecting it to a \(12.0 \mathrm{~V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\). (a) What are the angular frequency \(\omega\) of the electrical os cillations and the period of these ascillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) Ilow much energy is initially stored in the capacitor? (d) What is the charge on the capacitor \(0.0230 \mathrm{~s}\) after the connection to the inductor is made? Interpret the sign of your answer. (c) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

(a) What would have to be the self-inductance of a solenoid for it to store \(10.0 \mathrm{~J}\) of energy when a \(2.00 \mathrm{~A}\) current runs through it? (b) If this solenoid's cross-sectional diameter is \(4.00 \mathrm{~cm}\), and if you could wrap its coils to a density of 10 coils \(/ \mathrm{mm}\), how long would the solenoid be? (See Exercise \(30.11 .)\) Is this a realistic length for ordinary laboratory use?
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