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(a) A square loop of wire with sides of length \(40 \mathrm{~cm}\) is in a uniform magnetic field perpendicular to its area. If the field's strength is initially \(100 \mathrm{mT}\) and it decays to zero in \(0.010 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop? (b) What would be the average emf if the sides of the loop were only \(20 \mathrm{~cm} ?\)

Short Answer

Expert verified
(a) 1.6 V (b) 0.4 V

Step by step solution

01

Calculate the initial magnetic flux

The magnetic flux \( \Phi \) through one loop is given by the formula \( \Phi = B \cdot A \), where \( B \) is the magnetic field strength and \( A \) is the area of the loop. For a square loop, the area \( A \) is \( l^2 \) where \( l \) is the side length.For the initial magnetic field, \( B = 100 \ \text{mT} = 0.1 \ \text{T} \) and the side length is \( 40 \ \text{cm} = 0.4 \ \text{m} \). Therefore, the initial magnetic flux is:\[ \Phi_i = B \cdot A = 0.1 \ \text{T} \times (0.4 \ \text{m})^2 = 0.1 \times 0.16 = 0.016 \ \text{Wb} \]
02

Calculate the final magnetic flux

At the final time \( t = 0.010 \ \text{s} \), the magnetic field has decayed to zero. Therefore, the final magnetic flux \( \Phi_f \) is:\[ \Phi_f = B_f \cdot A = 0 \cdot (0.4 \ \text{m})^2 = 0 \ \text{Wb} \]
03

Calculate the change in magnetic flux

The change in magnetic flux \( \Delta \Phi \) is given by:\[ \Delta \Phi = \Phi_f - \Phi_i = 0 - 0.016 \ = -0.016 \ \text{Wb} \]
04

Calculate the average emf for a 40 cm side loop

According to Faraday's law of electromagnetic induction, the average emf \( \varepsilon \) induced in a loop is given by the rate of change of magnetic flux:\[ \varepsilon = -\frac{\Delta \Phi}{\Delta t} \]Given \( \Delta \Phi = -0.016 \ \text{Wb} \) and \( \Delta t = 0.010 \ \text{s} \), the magnitude of the average emf is:\[ \varepsilon = -\frac{-0.016}{0.010} = 1.6 \ \text{V} \]
05

Repeat Steps 1-4 for a 20 cm side loop

We will calculate the equivalent values for a 20 cm loop:- **New Area**: \( A = (0.2)^2 = 0.04 \ \text{m}^2 \)- **Initial Flux**: \( \Phi_i = 0.1 \times 0.04 = 0.004 \ \text{Wb} \)- **Final Flux**: \( \Phi_f = 0 \ \text{Wb} \)- **Change in Flux**: \( \Delta \Phi = -0.004 \ \text{Wb} \)- **Average EMF**: \( \varepsilon = \frac{-0.004}{0.010} = 0.4 \ \text{V} \)

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Key Concepts

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

Magnetic Flux
Understanding magnetic flux is crucial for grasping electromagnetic induction. Magnetic flux, denoted by the Greek letter \( \Phi \), is a measure of the quantity of magnetism, taking account of the strength and the extent of a magnetic field. Imagine it as the number of magnetic field lines passing through a certain area.

The formula for magnetic flux is:
  • \( \Phi = B \cdot A \cdot \cos(\theta) \)
where:
  • \( B \) is the magnetic field strength
  • \( A \) is the area through which the field lines pass
  • \( \theta \) is the angle between the magnetic field and the perpendicular to the area
In our scenario, the magnetic field is perpendicular to the loop, so \( \theta = 0 \) and \( \cos(\theta) = 1 \). Thus, the magnetic flux simplifies to \( \Phi = B \cdot A \).
For the exercise, with a square loop where sides are 40 cm, initially \( B = 0.1 \text{ T} \), and \( A = 0.16 \text{ m}^2 \), so \( \Phi_i = 0.016 \text{ Wb} \). As the field diminishes to zero, the magnetic flux also falls to zero.
Faraday's Law
Michael Faraday discovered that a change in magnetic flux can induce an electromotive force (emf) within a loop, leading to a current if the circuit is closed. This principle is encapsulated in Faraday's Law of Electromagnetic Induction. Faraday’s law can be written as:
  • \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \)
where:
  • \( \varepsilon \) is the induced emf
  • \( \Delta \Phi \) is the change in magnetic flux
  • \( \Delta t \) is the time interval over which this change occurs
The negative sign indicates the direction of the induced emf as per Lenz's Law. This law states that the direction of induced current is such that it opposes the change in magnetic flux that produced it.

In our exercise, the magnetic flux changes from 0.016 Wb to 0 Wb within 0.010 s, so the rate of change of flux is significant enough to induce a measurable emf.
Emf Calculation
Calculating the electromotive force (emf) effectively demonstrates Faraday's Law in practical scenarios. For a loop of wire, when the magnetic field through it changes, an emf is induced according to the rate of change of the magnetic flux.

In our given exercise:- For the 40 cm loop: We have \( \Delta \Phi = -0.016 \text{ Wb} \) and the time \( \Delta t = 0.010 \text{ s} \). Plugging into Faraday’s law gives \( \varepsilon = \frac{0.016}{0.010} = 1.6 \text{ V} \).- For the 20 cm loop, the same process applies, with a smaller area resulting in smaller \( \Delta \Phi = -0.004 \text{ Wb} \). The induced emf here is \( \varepsilon = \frac{0.004}{0.010} = 0.4 \text{ V} \).The key takeaway is that the larger the area and the faster the change, the greater the induced emf will be. This understanding is vital in designing electrical devices like transformers and generators, where magnetic flux changes are a core operational principle.

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Most popular questions from this chapter

A student makes a simple ac generator by using a single square wire loop \(10 \mathrm{~cm}\) on a side. The loop is then rotated at a frequency of \(60 \mathrm{~Hz}\) in a magnetic field of \(0.015 \mathrm{~T}\). (a) What is the maximum emf output? (b) If she wanted to make the maximum emf output ten times larger by adding loops, how many should she use in total?

When the magnetic flux through a single loop of wire increases by \(30 \mathrm{~T} \cdot \mathrm{m}^{2},\) an average current of \(40 \mathrm{~A}\) is induced in the wire. Assuming that the wire has a resistance of \(2.5 \Omega,\) (a) over what period of time did the flux increase? (b) If the current had been only \(20 \mathrm{~A},\) how long would the flux increase have taken?

A solenoid of length \(40.0 \mathrm{~cm}\) is made of 10000 circular coils. It carries a steady current of 12.0 A. Near its center is placed a small, flat, circular metallic coil of 200 circular loops, each with a radius of \(2.00 \mathrm{~mm}\). This small coil is oriented so that it receives half of the maximum magnetic flux. A switch is opened in the solenoid circuit and its current drops to zero in \(25.0 \mathrm{~ms}\). (a) What was the initial flux through the small coil? (b) Determine the average induced emf in the small coil during the \(25.0 \mathrm{~ms}\). (c) If you look along the long axis of the solenoid so that the initial 12.0 A current is clockwise, determine the direction of the induced current in the small inner coil during the time the current drops to zero. (d) During the \(25.0 \mathrm{~ms}\), what was the average current in the small coil, assuming it has a resistance of \(0.15 \Omega ?\)

In \(0.20 \mathrm{~s}\), a coil of wire with 50 loops experiences an average induced emf of \(9.0 \mathrm{~V}\) due to a changing magnetic field perpendicular to the plane of the coil. The radius of the coil is \(10 \mathrm{~cm}\), and the initial strength of the magnetic field is 1.5 T. Assuming that the strength of the field decreased with time, (a) what is the final strength of the field? (b) If the field strength had, instead, increased, what would its final value have been? (c) Explain a method whereby you could, in principle, tell whether the field was increasing or decreasing in magnitude.

The transformer on a utility pole steps the voltage down from \(10000 \mathrm{~V}\) to \(220 \mathrm{~V}\) for use in a college science building. During the day, the transformer delivers electric energy at the rate of \(10.0 \mathrm{~kW}\). (a) Assuming the transformer to be ideal, during that time, what are the primary and secondary currents in the transformer? (b) If the transformer is only \(90 \%\) efficient (but still delivers electric power at \(10.0 \mathrm{~kW}\) ), how does its inputcurrent compare to the ideal case? (c) At what rate is heat lost in the nonideal transformer? (d) If you wanted to keep the transformer \(\mathrm{cool}\) and to do this needed to dissipate half of the joule heating of part (c) using water cooling lines (the other half is taken care of by air cooling), what should be the rate of flow (in liters per minute) of water in the lines? Assume the input cool water is at \(68^{\circ} \mathrm{F}\) and the maximum allowable output water temperature is \(98^{\circ} \mathrm{F}\).

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