/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 A solenoidal coil with 25 turns ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The average magnetic flux through each turn of the inner solenoid is \(0.000706 \, \mathrm{T \cdot m^{2}}\). The mutual inductance of the two solenoids is \(0.176 \, \mathrm{H}\). The emf induced in the outer solenoid by the changing current in the inner solenoid is \(0.264 \, \mathrm{V}\).

Step by step solution

01

Calculating the Magnetic Flux

Calculating the magnetic flux through each turn of the inner solenoid involves using the formula for the magnetic field of a solenoid \(B = \mu_{0} n I\), where \(\mu_{0}\) is the permeability of free space, \(n\) is the number of turns per unit length, and \(I\) is the current. To find \(n\), divide the total number of turns by the length of the solenoid. Multiply \(B\) by the cross-sectional area of the solenoid to get the magnetic flux per turn.
02

Finding the mutual inductance

The mutual inductance \(M\) of the two solenoids can be calculated using the formula \(M = N\phi / I\), where \(N\) is the total number of turns of the outer solenoid, \(\phi\) is the magnetic flux through each turn calculated in previous step and \(I\) is the current in the inner solenoid.
03

Computing the induced emf

The emf induced in the outer solenoid by the changing current in the inner solenoid can be calculated using Faraday's law in the form \(\Delta V = -M\Delta I/\Delta t\), where \(\Delta I/\Delta t\) is the rate of change of the current in the inner solenoid, and \(M\) is the mutual inductance calculated in Step 2.

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Key Concepts

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

Solenoid
A solenoid is a type of electromagnet, essentially a coil of wire, that is designed to produce a controllable magnetic field. The magnetic field strength within a solenoid is directly proportional to the number of turns in the coil and the electric current passing through it. To visualize it, imagine wrapping a wire around a pencil; each loop of the wire is a turn, and a solenoid is just many such turns aligned along a common axis.

When an electrical current flows through this coil, a magnetic field is produced inside. The strength of this magnetic field can be determined by the formula:
\[ B = \frac{\text{number of turns per unit length} \times \text{magnetic permeability of free space} \times \text{current}}{ \text{length}} \]
In our exercise, the solenoid has 25 turns and carries a current, creating a magnetic field within its confines. Understanding the properties and behavior of a solenoid is critical for applications like electromagnets, inductors in electronic circuits, and devices like MRI machines.
Mutual Inductance
Mutual inductance is a principle that describes how two inductors exchange energy through their mutual magnetic fields. It quantifies the ability of one circuit to induce an electromotive force (EMF) in another through electromagnetic induction when the current in the first circuit changes.

The mutual inductance, denoted as \( M \), is given by the equation:
\[ M = \frac{N\text{φ}}{I} \]
where \( N \) is the number of turns in the secondary coil, \( \text{φ} \) is the magnetic flux through one turn of the primary coil, and \( I \) is the current in the primary coil. In the provided exercise, the mutual inductance is calculated to find out how the changing current in the inner solenoid affects the outer solenoid. This concept is highly relevant in the design of transformers and various wireless communication technologies.
Faraday's Law
Faraday's Law is a fundamental principle of electromagnetism which describes how a change in magnetic field can induce an electromotive force (EMF) in a coil of wire. The law states that the induced EMF is equal to the rate at which the magnetic flux through the circuit changes. Mathematically, the law is expressed as:
\[ \text{Induced EMF} = -\frac{\text{dφ}}{\text{dt}} \]
where the negative sign indicates the direction of the induced EMF (as per Lenz's Law), which opposes the change in flux that produced it. In the context of our problem, Faraday’s Law is used to calculate the EMF induced in the outer solenoid when the current in the inner solenoid changes. Understanding Faraday's Law is essential for grasping how electric generators, transformers, and induction motors operate.
Electromagnetic Induction
Electromagnetic induction is the process by which a conductor placed in a changing magnetic field (or a conductor moving through a stationary magnetic field) causes the production of a voltage across the conductor. This phenomenon was discovered by Michael Faraday in the 19th century, forming the basis for many electrical technologies.

Electromagnetic induction is responsible for the induced EMF in devices like generators, where a coil of wire rotates in a magnetic field, and inductively coupled devices such as wireless charging stations, where a changing magnetic field in one coil induces a current in another. In our textbook problem, the changing current in the inner solenoid induces an EMF in the outer solenoid, illustrating the direct application of this fundamental concept. Understanding electromagnetic induction is vital for studying how electrical energy can be generated and manipulated without physical connections.

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Most popular questions from this chapter

Two coils have mutual inductance \(M=3.25 \times 10^{4}\) H. The current \(i_{1}\) in the first coil increases at a uniform rate of \(830 \mathrm{~A} / \mathrm{s}\). (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

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

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 \(I=\frac{1}{2} M a^{2}\).

Inductance of a Solenoid. (a) A long, straight solenoid has \(N\) turns, uniform cross-sectional area \(A,\) and length \(1 .\) Show that the inductance of this solenoid is given by the equation \(L=\mu_{0} A N^{2} / l\) Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) \(A\) metallic laboratory spring is typically \(5.00 \mathrm{~cm}\) long and \(0.150 \mathrm{~cm}\) in diameter and has 50 coils. If you connect such a spring in an clectric circuit, how much self-inductance must you include for it if you model it as an idcal solcnoid?

A long. straight solenoid has 800 turns. When the current in the solenoid is \(2.90 \mathrm{~A}\), the average flux through each turn of the solenoid is \(3.25 \times 10^{-3} \mathrm{~Wb}\). What must be the magnitude of the rate of change of the current in order for the sclf-induced cmf to cqual \(6.20 \mathrm{mV} ?\)

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