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A 2.50-mH toroidal solenoid has an average radius of 6.00 cm and a cross- sectional area of 2.00 cm2. (a) How many coils does it have? (Make the same assumption as in Example 30.3.) (b) At what rate must the current through it change so that a potential difference of 2.00 V is developed across its ends?

Short Answer

Expert verified
The toroidal solenoid has approximately 239 coils, and the current must change at 800 A/s to achieve a 2.00 V potential difference.

Step by step solution

01

Convert Units to Standard Units

Convert the given values:- The inductance L=2.50mH=2.50×103H.- The average radius of the toroid r=6.00cm=0.0600m.- The cross-sectional area A=2.00cm2=2.00×104m2.
02

Use the Formula for Inductance of a Toroid

The formula for the inductance of a toroidal solenoid is given by:L=μ0N2A2πrWhere μ0 is the permeability of free space 4π×107H/m. Rearrange this equation to solve for N, the number of coils:N=2πrLμ0A
03

Substitute Known Values for N

Substitute the known values into the rearranged formula:N=2π(0.0600)(2.50×103)4π×107(2.00×104)After calculating, you find:N239So, the number of coils is approximately 239.
04

Using Faraday's Law of Induction

To find the rate of change of current, use Faraday's law of induction, which relates induced electromotive force (emf, ε) to the rate of change of current:ε=LdIdtGiven ε=2.00V, rearrange to solve for dIdt:dIdt=εL
05

Substitute Values into Faraday's Law Equation

Insert the known values into the equation for the rate of change of current:dIdt=2.002.50×103Calculate to find:dIdt800A/sThis means the current must change at a rate of 800 A/s to develop a potential difference of 2.00 V across the solenoid.

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Key Concepts

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

Toroidal Solenoid
A toroidal solenoid is a coil of wire that is shaped like a donut or a ring. This shape makes it quite different from the more common, straight solenoids. One of the key properties of a toroidal solenoid is that the magnetic field it generates stays within the body of the shape.
This feature helps reduce the magnetic field outside the solenoid and is utilized in many practical applications. When it comes to figuring out how many turns or coils a toroidal solenoid has, you generally use the formula for its inductance, which involves the permeability of free space, the solenoid's cross-sectional area, and its average radius. This allows us to calculate a number known as the inductance, which measures how well a solenoid can store energy in its magnetic field. Inductance is denoted by L and for a toroidal solenoid, the formula is:L=μ0N2A2πrHere, N is the number of coils, A represents the cross-sectional area, and r is the average radius of the toroid. Knowing how to rearrange and plug in the values for these variables helps you find the number of coils needed for your specific requirements.
Faraday's Law of Induction
Faraday's Law of Induction is a fundamental principle that explains how electrical currents are produced by changing magnetic fields. It highlights the relationship between electromotive force (emf) and the change in magnetic flux. This law is at the heart of many electrical devices, including generators and transformers.
The law states that the induced emf in a closed loop is equal to the negative rate of change of the magnetic flux through the loop. Mathematically, it's represented as:ε=dΦdtwhere ε is the emf, and dΦdt is the rate of change of magnetic flux. In cases involving inductance, this law is often written using the concept of inductance and current, given as:ε=LdIdtHere, L is the inductance and dIdt is the rate of change of current.
This equation tells us that any change in the magnetic environment of the coil will induce an emf that opposes this change. Hence, Faraday's Law sheds light on how effectively inductors respond to changes in current. This is crucial for the design and functioning of electric circuits and many electronic devices.
Electromotive Force (emf)
Electromotive Force, or emf, is a term used to describe the potential difference generated by a change in magnetic flux. Although it sounds like a force, it is actually a potential difference, similar to voltage, that results from an energy conversion process. In electrical devices, emf is critical as it causes current to flow when a circuit is complete.
In the context of a toroidal solenoid, the emf is produced when there is a change in the current flowing through the solenoid. Using Faraday's Law, we can understand how variations in current lead to an emf being induced. The formula to calculate emf in such cases is essentially:ε=LdIdtHere, a negative sign indicates that the induced emf works against the change in current, following Lenz's law that highlights nature's resistance to changes in magnetic fields.
The calculation involving emf becomes significant when designing circuits that require specific current rates to maintain efficient functionality. In practical applications, understanding and controlling emf helps enhance the reliability and safety of electronic systems.

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Most popular questions from this chapter

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection is completed. After a long time, the current is 6.45 mA. What are (a) the resistance R of the inductor and (b) the inductance L of the inductor?

A 7.50-nF capacitor is charged up to 12.0 V, then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be 8.60 × 105 s. Calculate: (a) the inductance of the coil; (b) the maximum charge on the capacitor; (c) the total energy of the circuit; (d) the maximum current in the circuit.

An inductor with an inductance of 2.50 H and a resistance of 8.00 Ω 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.

A 35.0-V battery with negligible internal resistance, a 50.0-Ω resistor, and a 1.25-mH inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach one-half of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

At the instant when the current in an inductor is increasing at a rate of 0.0640 A/s, the magnitude of the self-induced emf is 0.0160 V. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 A?

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