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A 100-turn, 4.0-cm-diameter coil has a resistance of 1.0 \Omega. A magnetic ficld perpendicular to the coil is \(B=t-\frac{1}{4} t^{2},\) where \(B\) is in tesla and \(t\) is in seconds. a. Draw a graph of \(B\) as a function of time from \(t=0\) s to \(t=4 \mathrm{s}\) b. Find an expression for the induced current \(I(t)\) as a function of time. c. Evaluate \(I\) at \(t=1,2,\) and 3 s.

Short Answer

Expert verified
The induced current \(I(t)\) as a function of time is given by \(I(t) = -N*(1 - \frac{1}{2}t)\). At \(t=1\) sec, \(I= -50 A\), at \(t=2\) sec, \(I = 0 A\) and at \(t=3\) sec, \(I = 50 A\).

Step by step solution

01

Graph the magnetic field as a function of time

Plot the function \(B(t) = t - \frac{1}{4}t^2\) on a graph with the x-axis representing time (0 to 4 sec) and the y-axis representing the magnetic field strength. The graph shows the magnetic field strength decreasing as time increases.
02

Derive the expression for induced current

The induced electromotive force (emf) can be calculated using Faraday's Law of electromagnetic induction, which states that emf = -N*dB/dt, where N is the number of turns in the coil and dB/dt is the rate of change of the magnetic field. The induced current is then determined by Ohm's Law, which states that current, I = emf / R, where R is the resistance of the coil. Hence, differentiate the given magnetic field equation with respect to time to achieve \(\frac{dB}{dt} = 1 - \frac{1}{2}t \). Substitute N = 100, and R = 1 Ω, we derive \(I(t) = -N*(1 - \frac{1}{2}t)\)
03

Evaluate the induced current at specific times

To find the current at \(t=1, 2, 3\) s, simply substitute these values into the equation derived in step 2. Hence, \(I(1) = -100*(1 - \frac{1}{2}*1) = -50 A, I(2) = -100*(1 - \frac{1}{2}*2) = 0 A, I(3) = -100*(1 - \frac{1}{2}*3) = 50 A\)

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Key Concepts

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

Faraday's Law of Electromagnetic Induction
Imagine discovering a way to harness electricity just through the magic of moving magnets and coils. That's what Michael Faraday did back in the 19th century with his groundbreaking Faraday's Law of Electromagnetic Induction. Simply put, this law reveals that a voltage is induced in a circuit whenever it is exposed to a changing magnetic field. This is the principle behind many of today's electrical generators.

The mathematical beauty of Faraday's Law is found in the equation emf = -N * (dB/dt), where emf stands for electromotive force, N is the number of turns in the coil, and dB/dt represents the rate of change of the magnetic field. The negative sign in Faraday's Law is a nod to Lenz's Law, which tells us that the induced emf will always work to create a current whose magnetic field opposes the change that produced it—nature's way of saying 'you can't get anything for nothing'.
Induced Electromotive Force (emf)
Now let’s switch gears and think about this buzz word: electromotive force, or emf. It's not actually a force, despite the name—it's the potential difference or voltage induced by changing a magnetic field. Think of it as the push or the oomph that gets charges moving in a circuit to create an electric current.

This induced emf can be temporary, just the initial kickstart needed to get electrons on their way, occurring only while the magnetic field is changing. In our exercise, we calculated the induced emf by differentiating the magnetic field over time and multiplying by the number of turns in the coil. And as we can see, even if emf is a misnomer, the 'force' it implies is mighty enough to create electricity out of motion!
Ohm's Law
Quite handy in the electrical world, Ohm's Law is the bread and butter for understanding how current flows. It tells us how voltage, current, and resistance play together in a circuit. Ohm's Law states that the current I through a conductor between two points is directly proportional to the voltage V across the two points and inversely proportional to the resistance R of the conductor. The formula is often written as I = V/R.

In the context of our coil and its changing magnetic field, Ohm's Law helps us turn the induced emf into actual current. We figure out the push (emf) and then see how much that push flows through the coil's resistance to create an induced current. It's like knowing how much water will flow through a pipe if we know the water pressure and how thick the pipe is.
Magnetic Field Over Time
As the name suggests, 'magnetic field over time' pertains to how the strength of a magnetic field changes as time passes. This can be visualized through a graph that plots the magnetic field strength on the y-axis against time on the x-axis. The slope of this graph at any given point is actually the 'rate of change of the magnetic field', which is a key component in Faraday's Law.

In our exercise, we plotted the given formula for the magnetic field B = t - 1/4t^2 against time. The resulting graph helped us intuitively understand how the magnetic field increased at first but then began to decrease, which is crucial for predicting how it would induce an emf in our coil.
Rate of Change of Magnetic Field
The 'rate of change of magnetic field' is essentially the speed at which the magnetic field is getting stronger or weaker over time. It's the mathematical derivative of the magnetic field with respect to time, denoted as dB/dt. This concept is central to Faraday’s Law, as it determines the amount of emf induced in a coil. In straightforward terms, a faster changing magnetic field means a stronger induced emf.

During our exercise, we calculated this rate by differentiating the magnetic field formula. This told us exactly how quickly the magnetic field was changing at any given moment, so we knew how powerful the induced emf and subsequent current would be at those times, leading to the fascinating result that the current direction changes over time, indicative of how dynamic and responsive electrical systems are to their magnetic environments.

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Most popular questions from this chapter

A loop antenna, such as is used on a television to pick up UHF broadcasts, is 25 cm in diameter. The plane of the loop is perpendicular to the oscillating magnetic field of a \(150 \mathrm{MHz}\) electromagnetic wave. The magnetic field through the loop is \(B=(20 \mathrm{nT}) \sin \omega t\) a. What is the maximum emf induced in the antenna? b. What is the maximum emf if the loop is turned \(90^{\circ}\) to be perpendicular to the oscillating electric field?

A 2.0-cm-diameter solenoid is wrapped with 1000 turns per meter. \(0.50 \mathrm{cm}\) from the axis, the strength of an induced electric field is \(5.0 \times 10^{-4} \mathrm{V} / \mathrm{m} .\) What is the rate \(d I / d t\) with which the current through the solenoid is changing?

An \(8.0 \mathrm{cm} \times 8.0 \mathrm{cm}\) square loop is halfway into a magnetic field perpendicular to the plane of the loop. The loop's mass is \(10 \mathrm{g}\) and its resistance is \(0.010 \Omega .\) A switch is closed at \(t=0 \mathrm{s}\) causing the magnetic field to increase from 0 to \(1.0 \mathrm{T}\) in \(0.010 \mathrm{s}\) a. What is the induced current in the square loop? b. What is the force on the loop when the magnetic field is \(0.50 \mathrm{T} ?\) Is the force directed into the magnetic field or away from the magnetic field? c. What is the loop's acceleration at \(t=0.005\) s, when the field strength is 0.50 T T If this acceleration stayed constant, how far would the loop move in 0.010 s? d. Because \(0.50 \mathrm{T}\) is the average field strength, your answer to \(\mathrm{c}\) is an estimate of how far the loop moves during the \(0.010 \mathrm{s}\) in which the field increases to \(1.0 \mathrm{T}\). If your answer is \(<8 \mathrm{cm}\), then it is reasonable to neglect the movement of the loop during the 0.010 s that the field ramps up. Is neglecting the movement reasonable? e. With what speed is the loop "kicked" away from the magnetic field? Hint: What is the impulse on the loop?

Electricity is distributed from electrical substations to neighborhoods at \(15,000 \mathrm{V} .\) This is a \(60 \mathrm{Hz}\) oscillating \((\mathrm{AC})\) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the \(120 \mathrm{V}\) that is delivered to your house. a. How many turns does the primary coil on the transformer have if the secondary coil has 100 turns? b. No energy is lost in an ideal transformer, so the output power \(P_{\text {out }}\) from the secondary coil equals the input power \(P_{\text {in }}\) to the primary coil. Suppose a neighborhood transformer delivers \(250 \mathrm{A}\) at \(120 \mathrm{V} .\) What is the current in the \(15,000 \mathrm{V}\) line from the substation?

A circular loop made from a flexible, conducting wire is shrinking. Its radius as a function of time is \(r=r_{0} e^{-\beta t} .\) The loop is perpendicular to a steady, uniform magnetic field \(B\). Find an expression for the induced emf in the loop at time \(t.\)

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