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Chapter 30: Problem 8

When the currcnt in a toroidal solcnoid is changing at a rate of \(0.0260 \mathrm{~A} / \mathrm{s},\) the magnitude of the induced \(\mathrm{cmf}\) is \(12.6 \mathrm{mV}\). When the current equals \(1.40 \mathrm{~A},\) the average flux through each turn of the solenoid is 0.00285 Wb. Ilow many turns does the solenoid have?

Short Answer

Expert verified
The solenoid has approximately 175 turns.

Step by step solution

01

Write down the given values from the problem

From the problem, we know that the rate of change of current \(dI/dt = 0.0260 \, \mathrm{A/s}\), the induced emf \(\mathcal{E} = 12.6 \, \mathrm{mV} = 12.6 \times 10^{-3} \, \mathrm{V}\), and the average flux \(\Phi = 0.00285 \, \mathrm{Wb}\) when the current \(I = 1.40 \, \mathrm{A}\).
02

Apply Faraday's Law

Faraday's law states that the magnitude of the induced emf is equal to the rate of change of magnetic flux through the circuit. In a solenoid, the magnetic flux \(\Phi\) is equal to the product of the number of turns \(N\) and the magnetic field \(B\), therefore \(\Phi = NBA\). Since the area \(A\) is constant, we can differentiate both sides with respect to time and obtain \(d\Phi/dt = Nd(BA)/dt = NBA(dI/dt)\). From this, we can express the induced emf as \(\mathcal{E} = NBA(dI/dt)\). Solving for \(N\) gives \(N = \mathcal{E}/(BA(dI/dt))\).
03

Calculate the number of turns

Substituting the given values into the equation from Step 2 gives \(N = \mathcal{E}/(BA(dI/dt)) = (12.6 \times 10^{-3}) / ( (0.00285)(0.0260) )\). Evaluating this expression gives \(N \approx 175\).

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Key Concepts

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

Electromagnetic Induction
Electromagnetic induction is the process by which an electric current is produced in a conductor when it is exposed to a changing magnetic field. This phenomenon is the cornerstone upon which electric generators and transformers are built and is described by Faraday's Law of Induction. According to Faraday's Law, an electromotive force (emf) is induced in a circuit whenever the magnetic flux linked with that circuit changes.

It's simple when you look at it this way: imagine you have a loop of wire in a magnetic environment. If you move the magnet or the wire, you're changing how much magnetic 鈥榮tuff' 鈥� known as flux 鈥� goes through the loop. A changing magnetic flux is what generates an electric current in the wire.

Moreover, the strength of the induced emf is directly proportional to the rate at which the magnetic flux changes through the loop. This is why in the textbook exercise, the induced emf is related to both the change in current and the magnetic properties of the solenoid 鈥� the toroidal shape that directs the magnetic field.
Solenoid
A solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field. If you've ever seen a coil of wire wound into a tightly packed helix, then you've seen a solenoid. When an electric current passes through the solenoid, the coil becomes magnetized, with distinct north and south poles forming, similar to a bar magnet.

To understand the operation of a solenoid, think of it as many small loops, called turns, stacked alongside each other. Each of these turns contributes a small amount of magnetic field, and when added together, they produce a significant and uniform field inside the coil.

In the exercise, the solenoid's turns are crucial because the magnetic flux that we discuss is the flux through one turn multiplied by the number of turns. More turns mean more opportunities for inducing emf when a current changes.
Magnetic Flux
Magnetic flux is, in simple terms, a measure of the quantity of the magnetic field that passes through a given area. It is denoted typically as \( \Phi \) and is a product of the field strength, the area it penetrates, and the angle between the magnetic field and the perpendicular to the surface.

The concept can be tricky, but you can visualize it by imagining a surface in a magnetic field. The number of magnetic field lines that pass through that surface represents the magnetic flux. Factors that can change the flux include varying the strength of the magnetic field, modifying the size of the surface, and altering the angle of the field relative to the surface.

In our exercise, we're looking at how the magnetic flux changes as the current in the solenoid changes. This change is what leads to the induction of an emf, and calculating the flux allows us to determine how many turns the solenoid has, by connecting the dots between the magnetic flux, the induced emf, and the rate of change of current.

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Most popular questions from this chapter

An inductor with an inductance of \(2.50 \mathrm{II}\) and a resistance of \(8.00 \Omega\) is connected to the terminals of a battery with an emf of \(6.00 \mathrm{~V}\) 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 \(0.500 \mathrm{~A}\) : (c) the current \(0.250 \mathrm{~s}\) after the circuit is closed; (d) the final steady-state current.

It is possible to make your own inductor by winding wire around a cylinder, such as a pcncil. Assume you have a spool of AWG 20 copper wire, which has a diameter of \(0.812 \mathrm{~mm}\). (a) Estimate the diameter of a pencil. (b) Estimate how many times can you tightly wrap AWG 20 copper wire around a pencil to form a solenoid with a length of \(4.0 \mathrm{~cm}\). (c) Estimate the inductance of this solcnoid by assuming the magnetic field inside is constant. (d) If a current of 1.0 A flows through this solenoid, how much magnetic energy will be stored inside?

CP An altemating-current clectric motor includes a thin. hollow, cylindrical spool (similar to a ring) with mass \(M=1.11 \mathrm{~kg}\) and radius \(a=5.00 \mathrm{~cm}\) wrapped \(N=500\) times with a copper wire with resistance \(R=5.00 \Omega\) and inductance \(L=77.0 \mathrm{mH}\). Within the spool is a battery that supplics current \(I=1.00 \mathrm{~A}\). which makes the spool a magnetic dipole with dipole moment \(\overrightarrow{\boldsymbol{\mu}}\) parallel to the cylinder axis. A constant magnetic field with magnitude \(B=2.00 \mathrm{~T}\) is supplied by an extemal stator magnet, while the spool turns freely on an axis perpendicular to its own axis. At a certain time, a bar is inserted, stopping the spool's motion (Fig. \(\mathrm{P} 30.51\) ). At that instant the angle between the spool axis and the magnetic field is \(\theta=45^{\circ}\). (a) What is the magnitude of the downward force \(\vec{F}\) applicd by the bar onto the spool immediatcly after the bar is inserted? (b) Later, at time \(t=0\) with spool still at rest, the cril is short-circuited and a constant counter- toryue \(\tau=0.500 \mathrm{~N} \cdot \mathrm{m}\) is applied. The current subsides, and the magnetic torque decreases exponentially. At what time \(t\) does the force applied by the bar vanish'? (Hint: Determine when the magnetic torque balances the counter-torque.) (c) After the spool rotatcs \(180^{\circ}\) it hecomes stuck on the top side of the bar. The counter-torque is no longer applied, and the switch is returned to its original position. After a long time, the bar is removed. What is the angular acceleration of the spool immediately after the bar is removed? The moment of incrtia of the spool for an axis along its diamcter is \(I=\frac{1}{2} M a^{2}\).

A toroidal solenoid has 500 turns, cross-sectional area \(6.25 \mathrm{~cm}^{2}\) and mean radius \(4.00 \mathrm{~cm}\). (a) Calculate the coil's self-inductance. (b) If the current decreases uniformly from \(5.00 \mathrm{~A}\) to \(2.00 \mathrm{~A}\) in \(3.00 \mathrm{~ms},\) calculate the sclf- induced emf in the coil. (c) The current is dirccted from terminal \(a\) of the coil to terminal \(b .\) Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a\) ? $

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1 ). The inner solenoid is \(25.0 \mathrm{~cm}\) long and has a diameter of \(2.00 \mathrm{~cm}\). At a certain time, the current in the inner solenoid is \(0.120 \mathrm{~A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{~A} / \mathrm{s}\). For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.
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