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Chapter 29: Problem 51

CALC A very long, straight solenoid with a cross-sectional area of 2.00 \(\mathrm{cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0,\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{A} / \mathrm{s}^{2}\right) t^{2}\) . A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 \(\mathrm{A}\) ?

Short Answer

Expert verified
The induced emf is 0.0182 V.

Step by step solution

01

Calculate the rate of change of current

The current in the primary solenoid is given by the equation \( i(t) = 0.160 \times t^2 \) A/s虏. First, differentiate \( i(t) \) with respect to time to find \( \frac{di}{dt} \). Differentiating, we have: \[ \frac{di}{dt} = 0.160 \times \frac{d}{dt}(t^2) = 0.320 \times t \text{ A/s}. \] We need \( t \) such that \( i(t) = 3.20 \text{ A} \). Thus, solve \( 0.160 \times t^2 = 3.20 \) to find \( t = \sqrt{\frac{3.20}{0.160}} = 5 \text{ s} \). Substitute \( t = 5 \text{ s} \) into \( \frac{di}{dt} = 0.320 \times t \) to get \( \frac{di}{dt} = 0.320 \times 5 = 1.6 \text{ A/s} \).
02

Calculate the magnetic flux linkage

The magnetic flux through the solenoid, \( \Phi_B \), is given by \( \Phi_B = B \cdot A \), where \( B \) is the magnetic field inside the solenoid \( B = \mu_0 \cdot n \cdot i \), \( A \) is the cross-sectional area, and \( n \) is the number of turns per unit length. Here, \( \mu_0 = 4\pi \times 10^{-7} \mathrm{T}\,\mathrm{m}/\mathrm{A} \), \( n = 90 \times 10^2 \) turns/m, and \( A = 2.00 \times 10^{-4} \mathrm{m}^2 \). Substituting, \[ \Phi_B = (4\pi \times 10^{-7} \times 9000 \cdot 3.20) \times (2 \times 10^{-4}) \]. Simplify to get \( \Phi_B \approx 7.25 \times 10^{-4} \mathrm{Wb} \).
03

Calculate the emf induced in the secondary winding

The induced emf in the secondary winding can be calculated using the equation \( \text{emf} = -N \frac{d\Phi_B}{dt} \), where \( N = 5 \) is the number of turns in the secondary winding. Since \( \Phi_B = B \times A \) and \( B \) is proportional to \( i \), then \( \frac{d\Phi_B}{dt} = A \cdot \mu_0 \cdot n \cdot \frac{di}{dt} \). Substitute values: \( \frac{d\Phi_B}{dt} = 2.00 \times 10^{-4} \times 4\pi \times 10^{-7} \times 9000 \times 1.6 \). This gives \( \frac{d\Phi_B}{dt} \approx 3.63 \times 10^{-3} \mathrm{Wb/s} \). Thus, \( \text{emf} = -5 \times 3.63 \times 10^{-3} \approx -0.0182 \text{ V} \).
04

Conclusion

The magnitude of the emf induced in the secondary winding at the instant the current in the solenoid is 3.20 A is approximately 0.0182 V.

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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 coil made by winding a wire in the shape of a helix. Solenoids are used in electromagnetism to produce a magnetic field. When you send an electric current through the wire of a solenoid, it creates a magnetic field around it. This field resembles that of a bar magnet with a clear north and south pole.

The strength of the magnetic field inside a solenoid depends on a few key factors:
  • The amount of current flowing through the wire
  • The number of loops (or turns) of the wire per unit length
  • The material around which the wire is wrapped (the core)
In this exercise, we consider a solenoid with many turns of wire, tightly packed to maximize the magnetic field strength. We also see how changes in this current over time can affect the electromagnetic properties of the solenoid.
Electromagnetic Induction
Electromagnetic induction is a phenomenon where a changing magnetic field within a closed loop induces an electric current. This principle was first discovered by Michael Faraday in the 19th century. It's the fundamental physics behind generators and transformers.

When the magnetic field changes, it causes electrons in the wire to move, creating a current. This is similar to how a battery works, but instead of chemical reactions supplying the electron motion, it's the changing magnetic field doing the job. If you take a magnet and move it through or around a coil of wire, or alternatively move the coil through a magnetic field, you generate electricity.

In this problem, we are changing the current through the solenoid, which changes the magnetic field, resulting in an induced emf in the secondary winding.
Faraday's Law
Faraday's Law of electromagnetic induction quantifies how a change in a magnetic field can create an induced electromotive force (emf) in a closed circuit. It states that the induced emf is directly proportional to the rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

\[ ext{emf} = -N rac{d ext{桅}_B}{dt} \]

Here, \( N \) is the number of turns in the coil, and \( rac{d ext{桅}_B}{dt} \) is the rate of change of magnetic flux. The negative sign indicates the direction of the induced current opposes the change in magnetic flux, as per Lenz's Law.

In our exercise, this law helps us calculate the emf developed in the secondary winding due to the changing current in the solenoid.
Magnetic Flux
Magnetic flux (\( ext{桅}_B \)) represents the amount of magnetic field passing through a given area. It basically tells you how many magnetic field lines exist over a certain surface.

The formula for magnetic flux is:

\[ ext{桅}_B = B imes A \]

Where:
  • \( B \) is the magnetic field strength
  • \( A \) is the area perpendicular to the magnetic field
In this exercise, the magnetic flux is calculated using the magnetic field within the solenoid, which is itself dependent on the number of wire turns, the current, and the solenoid's cross-sectional area. Magnetic flux is a crucial part of understanding and calculating induced emf.
Rate of Change of Current
The rate of change of current refers to how quickly the current flowing through a circuit changes with time. It is represented as \( rac{di}{dt} \) in equations.

Understanding the rate of change of current is important for calculating the changing magnetic fields that lead to electromagnetic induction. In our problem, the current in the solenoid is given by \( i(t) = 0.160 t^2 \) A/s虏, and its time derivative provides the rate at which the current is changing.

Differentiating the current expression tells us how rapidly the current is increasing or decreasing. This rate directly influences the magnetic flux change and hence the induced electromotive force (emf) according to Faraday's Law. When the current increases at a higher rate, the magnetic field also changes more rapidly, leading to a stronger induced emf.

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Most popular questions from this chapter

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could operate using a generator but you have no magnets. The earth's magnetic field at your location is horizontal and has magnitude \(8.0 \times 10^{-5} \mathrm{T},\) and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 \(\mathrm{V}\) and estimate that you can rotate the coil at 30 \(\mathrm{rpm}\) by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum number of turns the coil can have is 2000 . (a) What area must the coil have? (b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you think this device is feasible? Explain.

A Magnetic Exercise Machine. You have designed a new type of exercise machine with an extremely simple mechanism (Fig. E29.28). A vertical bar of silver (chosen for its low resistivity and because it makes the machine look cool) with length \(L=3.0 \mathrm{m}\) is free to move left or right without friction on silver \(L=3.0 \mathrm{m}\) is free to move left or right without friction on silver rails. The entire apparatus is placed in a horizontal, uniform magnetic field of strength 0.25 \(\mathrm{T}\) . When you push the bar to the left or right, the bar's motion sets up a current in the circuit that includes the bar. The resistance of the bar and the rails can be neglected. The magnetic field exerts a force on the current-carrying bar, and this force opposes the bar's motion. The health benefit is from the exercise that you do in working against this force, (a) Your design goal is that the person doing the exercise is to do work at the rate of 25 watts when moving the bar at a steady 2.0 \(\mathrm{m} / \mathrm{s}\) . What should be the resistance \(R ?\) (b) You decide you want to be able to vary the power required from the person, to adapt the machine to the person's strength and fitness. If the power is to be increased to 50 \(\mathrm{W}\) by altering \(R\) while leaving the other design parameters constant, should \(R\) be increased or decreased? Calculate the value of \(R\) for 50 \(\mathrm{W} .\) (c) When you start to construct a prototype machine, you find it is difficult to produce a \(0.25-\mathrm{T}\) magnetic field over such a large area. If you decrease the length of the bar to 0.20 \(\mathrm{m}\) while leaving \(B, v,\) and \(R\) the same as in part (a), what will be the power required of the person?

CALC A Changing Magnetic Field. You are testing a new data-acquisition system. This system allows you to record a graph of the current in a circuit as a function of time. As part of the test, you are using a circuit made up of a 4.00 -cm-radius, 500 -turn coil of copper wire connected in series to a \(600-\Omega\) resistor. Copper has resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m},\) and the wire used for the coil has diameter 0.0300 \(\mathrm{mm}\) . You place the coil on a table that is tilted \(30.0^{\circ}\) from the horizontal and that lies between the poles of an electromagnet. The electromagnet generates a vertically upward magnetic field that is zero for \(t<0,\) equal to \((0.120 \mathrm{T}) \times\) \((1-\cos \pi t)\) for \(0 \leq t \leq 1.00 \mathrm{s},\) and equal to 0.240 T for \(t>1.00 \mathrm{s}\) . (a) Draw the graph that should be produced by your data-acquisition system. (This is a full-featured system, so the graph will include labels and numerical values on its axes.) (b) If you were looking vertically downward at the coil, would the current be flowing clockwise or counterclockwise?

The magnetic field within a long, straight solenoid with a circular cross section and radius \(R\) is increasing at a rate of \(d B / d t\) . (a) What is the rate of change of flux through a circle with radius \(r_{1}\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_{1}\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance \(r_{2}\).from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r=0\) to \(r=2 R\) (e) What is the magnitude of the induced emf in a circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is \(R ?\) (g) What is the induced emf if the radius in part (e) is 2R?

CALC A coil 4.00 \(\mathrm{cm}\) in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B=(0.0120 \mathrm{T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{T} / \mathrm{s}^{4}\right) t^{4} .\) The coil is connected to a \(600-\Omega\) resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time \(t=5.00 \mathrm{s?}\)
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