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Chapter 20: Problem 11

A wire loop of radius \(0.90 \mathrm{~m}\) lies so that an external magnetic field of magnitude \(0.30 \mathrm{~T}\) is perpendicular to the loop. The field reverses its direction, and its magnitude changes to \(0.20 \mathrm{~T}\) in \(1.5 \mathrm{~s}\). Find the magnitude of the average induced emf in the loop during this time.

Short Answer

Expert verified
To calculate the magnitude of the average induced emf, first calculate the initial and final magnetic flux, then find the absolute difference between the two (this is the change in magnetic flux). Apply Faraday's law of electromagnetic induction to find the induced emf by dividing the change in magnetic flux by the change in time. Substituting the given values and doing the mathematical calculations yields the magnitude of the average induced emf.

Step by step solution

01

Calculate Initial Magnetic Flux

The magnetic flux \(桅1\) for a circular loop is given by \(桅1 = B1*A\), where \(B1\) is the initial magnetic field and \(A\) is the area of the loop which can be calculated using the formula \(蟺r^2\). So, \ \(B1 = 0.30 T\), \(r = 0.90 m\) and \(A = 蟺*(0.90)^2 m^2\). So, we calculate \(桅1 = B1*A = 0.30 T * 蟺*(0.90 m)^2\).
02

Calculate Final Magnetic Flux

Similarly, we calculate the final magnetic flux \(桅2 = B2*A\), where \(B2 = 0.20 T\) is the final magnetic field and \(A\) is the area of the loop. Therefore, \(桅2 = B2*A = 0.20 T * 蟺*(0.90 m)^2\).
03

Calculate the change in Magnetic Flux

The change in magnetic flux \(螖桅\), is the absolute difference between the final magnetic flux and initial magnetic flux i.e., \(螖桅 = |桅2 - 桅1|\). Substituting \(桅1\) and \(桅2\) we get \(螖桅 = |0.20 T * 蟺*(0.90m)^2 - 0.30 T * 蟺*(0.90m)^2|\).
04

Apply Faraday's Law

According to Faraday's law of electromagnetic induction, the induced emf is given by the change in magnetic flux divided by the change in time. Therefore, the magnitude of average induced emf \(蔚\) is \(蔚 = 螖桅 / 螖t\), where \(螖t = 1.5 s\) is the time interval. Substituting for \(螖桅\) and \(螖t\) we get \(蔚 = (|0.20 T * 蟺*(0.90m)^2 - 0.30 T * 蟺*(0.90m)^2|)/1.5 s\)

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Key Concepts

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

Magnetic Flux
Magnetic flux is a concept that helps us understand how much magnetic field "flows" through a given area. Imagine you have a loop of wire and a magnetic field pointing perpendicularly through it. The amount of magnetic flux is essentially how many magnetic field lines pass through that loop.
If the field is strong and the loop is large, you will have a high magnetic flux. It is mathematically represented by the formula \( \Phi = B \cdot A \cdot \cos(\theta) \), where:
  • \( B \) is the magnetic field strength (in teslas),
  • \( A \) is the area the magnetic field passes through (in square meters),
  • \( \theta \) is the angle between the magnetic field lines and the normal (perpendicular) line to the surface.
In our exercise, since the magnetic field is perpendicular to the loop, \( \theta = 0 \) and \( \cos(\theta) = 1 \), simplifying our calculations for magnetic flux.
Faraday's Law
Faraday's Law of Electromagnetic Induction is a principle that tells us how a changing magnetic flux can induce an electromotive force (emf) in a circuit. Basically, if the magnetic environment of a coil (or loop) changes, this change in magnetic flux can generate electricity.
Faraday's law is quantified by the equation \( \varepsilon = - \frac{d\Phi}{dt} \), where:
  • \( \varepsilon \) is the induced emf,
  • \( d\Phi \) is the change in magnetic flux,
  • \( dt \) is the change in time.
The negative sign in the formula reflects Lenz's Law, which states that the induced emf will generate a current that opposes the change in magnetic flux. In our example problem, we calculated the change in magnetic flux and divided it by the time over which it occurred to find the induced emf using Faraday's law.
Induced EMF
The induced electromotive force (emf) is essentially a voltage that is generated in a loop or coil due to a changing magnetic field. Just like a battery can push electrons through a circuit, an induced emf does the same due to changes in magnetic conditions.
In the context of our exercise, the key points to understand about induced emf include:
  • It is directly proportional to the rate of change of magnetic flux through the loop.
  • If the magnetic flux changes rapidly, a higher induced emf is generated.
  • It can be calculated using Faraday's law, as shown: \( \varepsilon = \frac{\Delta \Phi}{\Delta t} \).
This fundamental concept shows how mechanical energy (such as moving magnets) can be converted into electrical energy, which is at the heart of many devices, from electric generators to transformers.

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Most popular questions from this chapter

A technician wraps wire around a tube of length \(36 \mathrm{~cm}\) having a diameter of \(8.0 \mathrm{~cm}\). When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 580 turns of wire. (a) Find the self-inductance of this solenoid. (b) If the current in this solenoid increases at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\), what is the selfinduced emf in the solenoid?

An aluminum ring of radius \(5.00 \mathrm{~cm}\) and resistance \(3.00 \times 10^{-4} \Omega\) is placed around the top of a long air-core solenoid with 1000 turns per meter and a smaller radius of \(3.00 \mathrm{~cm}\), as in Figure \(\mathrm{P} 20.64\). If the current in the sole noid is increasing at a constant rate of \(270 \mathrm{~A} / \mathrm{s}\), what is the induced current in the ring? Assume the magnetic field produced by the solenoid over the area at the end of the solenoid is one-half as strong as the field at the center of the solenoid. Assume also the solenoid produces a negligible field outside its cross-sectional area.

An \(R L\) circuit with \(L=3.00 \mathrm{H}\) and an \(R C\) circuit with \(C=\) \(3.00 \mu \mathrm{F}\) have the same time constant. If the two circuits have the same resistance \(R\), (a) what is the value of \(R\) and (b) what is this common time constant?

The rolling axle, \(1.50 \mathrm{~m}\) long, is pushed along horizontal rails at a constant speed \(v=3.00 \mathrm{~m} / \mathrm{s}\). A resistor \(R=0.400 \Omega\) is connected to the rails at points \(a\) and \(b_{1}\) directly opposite each other. (The wheels make good electrical contact with the rails, so the axle, rails. and \(R\) form a closed-loop circuit. The only significant resistance in the circuit is \(R\).) A uniform magnetic field \(B=0.800 \mathrm{~T}\) is directed vertically downward. (a) Find the induced current \(I\) in the resistor. (b) What horizontal force \(\overrightarrow{\mathbf{F}}\) is required to keep the axle rolling at constant speed? (c) Which end of the resistor, \(a\) or \(b\), is at the higher electric potential? (d) After the axle rolls past the resistor, does the current in \(R\) reverse direction?

The current in a coil changes from \(3.5 \mathrm{~A}\) to \(2.0 \mathrm{~A}\) in \(0.50 \mathrm{~s}\). If the average emf induced in the coil is \(12 \mathrm{mV}\), what is the self-inductance of the coil?
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