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A 5.0-cm-diameter coil has 20 turns and a resistance of \(0.50 \Omega .\) A magnetic field perpendicular to the coil is \(B=\) \(0.020 t+0.010 t^{2},\) 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 \(t=10 \mathrm{s}\) b. Find an expression for the induced current \(I(t)\) as a function of time. c. Evaluate \(I\) at \(t=5 \mathrm{s}\) and \(t=10 \mathrm{s}\).

Short Answer

Expert verified
After performing the steps, the graph of \(B(t)\) is obtained in step 1. Then, the general formula for \(I(t)\) is obtained in step 2 by applying Faraday's law of electromagnetic induction and Ohm's law. In step 3, the specific values of the current at \(t=5s\) and \(t=10s\) are calculated by substituting into the \(I(t)\) equation.

Step by step solution

01

Draw the graph of B as a function of time

The magnetic field B is given as a function of time t as \(B(t) = 0.020t + 0.010t^{2}\). Using any graph plotting tool, a graph can be plotted with t-axis from 0 to 10 seconds and B-axis according to the above equation.
02

Find the expression for induced current

According to Faraday's law of electromagnetic induction, the magnitude of the induced emf e in a coil is given by the rate of change of magnetic flux. Since flux \(\Phi = BAcos\theta\), where A is the area of coil, B is the magnetic field, and theta is the angle between them, for our problem, this simplifies to e = - \(\frac{d(BA)}{dt}\). The negative sign is due to Lenz's law and is not required for magnitude. Given that B = \(0.020t + 0.010t^{2}\), and A = \(\pi(0.025)^2\) (the coil is a circle with diameter 5.0 cm, so the radius is 2.5 cm or 0.025m), differentiate BA with respect to t to get emf. The current I can then be found by Ohm's law, \(e = I \cdot R\), solving for I, which is \(I = \frac{e}{R}\). The resistance R is given as 0.50 \(\Omega\).
03

Evaluate I at \(t=5s\) and \(t=10s\)

Substitute \(t=5s\) and \(t=10s\) into the I(t) equation obtained in step 2, which will yield the induced current at \(t=5s\) and \(t=10s\). These solve into specific values.

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Key Concepts

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

Faraday's Law
Faraday's Law of Electromagnetic Induction is a fundamental concept in physics that tells us how changing magnetic fields can induce an electromotive force (emf) in a coil or loop. This phenomenon is what makes electric generators possible by converting mechanical energy into electrical energy.

The key principle of Faraday’s Law is that the emf induced is proportional to the rate of change of the magnetic flux. Mathematically, this can be expressed as: \[ e = - \frac{d\Phi}{dt} \] where
  • \( e \) is the emf
  • \( \Phi \) is the magnetic flux
  • The negative sign represents Lenz’s Law, which indicates that the induced emf acts to oppose the change in flux.


In our exercise, the magnetic field changes over time, leading to a change in magnetic flux which then induces an emf and subsequently a current in the coil.
Induced Current
Induced current refers to the current generated in a circuit due to a changing magnetic field, as explained by Faraday’s Law. This current is generated because the changing magnetic environment causes electrons to move, creating a current.

In the given exercise, the magnetic field \( B(t) = 0.020t + 0.010t^{2} \) changes with time, affecting the coil. The emf is generated, which is given by the formula: \[ e = - \frac{d(BA)}{dt} \] The area \( A \) of the coil is calculated using the formula for the area of a circle \( A = \pi r^{2} \), where \( r = 0.025 \text{ m} \), the radius of the coil. Differentiating \( BA \) with respect to \( t \), we find the emf and subsequently the induced current \( I \): \[ I(t) = \frac{e}{R} \] where \( R \) is the resistance of the coil, given as \( 0.50 \Omega \). Then, using specific values of \( t \), you can calculate \( I \) at different times.
Magnetic Flux
Magnetic flux represents the total magnetic field passing through a given area. It is a crucial term in Faraday’s Law and is defined by the equation: \[ \Phi = B \cdot A \cdot \cos \theta \] where
  • \( B \) is the magnetic field
  • \( A \) is the area through which the field lines pass
  • \( \theta \) is the angle between the field lines and the normal to the surface


For the exercise at hand, since the field is perpendicular to the coil, the angle \( \theta \) is 0, and \( \cos 0 = 1 \), simplifying the calculation to \( \Phi = B \cdot A \). The magnetic flux is what directly leads to the induced emf when it changes over time, as Faraday’s Law specifies.
Ohm's Law
Ohm's Law is an essential principle for understanding the relationship between voltage, current, and resistance in an electrical circuit. It is represented by the equation: \[ V = I \cdot R \] In the context of electromagnetic induction, the emf \( (e) \) plays the role of \( V \) (voltage) in Ohm's Law. Therefore, for a coil with resistance \( R \), the induced current \( I \) can be determined as: \[ I = \frac{e}{R} \] This equation is particularly useful in the exercise since it allows us to find the induced current whenever we know the emf and the resistance.

By using the resistance of \( 0.50 \Omega \) and the emf calculated from Faraday's Law, you can easily determine the induced current at various points in time, such as \( t = 5s \) and \( t = 10s \). Ohm's Law connects the effects of electromagnetic induction to measurable electrical outputs.

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Most popular questions from this chapter

The magnetic field inside a 5.0 -cm-diameter solenoid is \(2.0 \mathrm{T}\) and decreasing at \(4.0 \mathrm{T} / \mathrm{s} .\) What is the electric field strength inside the solenoid at point (a) on the axis and (b) \(2.0 \mathrm{cm}\) from the axis?

A 1000-turn coil of wire 2.0 cm in diameter is in a magnetic field that drops from \(0.10 \mathrm{T}\) to \(0 \mathrm{T}\) in \(10 \mathrm{ms}\). The axis of the \(\mathrm{coil}\) is parallel to the field. What is the emf of the coil?

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?

A \(20 \mathrm{cm} \times 20 \mathrm{cm}\) square loop has a resistance of \(0.10 \Omega . \mathrm{A}\) magnetic field perpendicular to the loop is \(B=4 t-2 t^{2},\) where \(B\) is in tesla and \(t\) is in seconds. a. Determine \(B, \mathcal{E},\) and \(I\) at half-second intervals from 0 s to 2 s. b. Use your results of part a to draw graphs of \(B\) and \(I\) versus time.

Your camping buddy has an idea for a light to go inside your He happens to have a powerful (and heavy!) horseshoe magnet that he bought at a surplus store. This magnet creates a \(0.20 \mathrm{T}\) field between two pole tips \(10 \mathrm{cm}\) apart. His idea is to build a hand-cranked generator with a rotating 5.0-cm-radius semicircle between the pole tips. He thinks you can make enough current to fully light a \(1.0 \Omega\) lightbulb rated at 4.0 W. That's not super bright, but it should be plenty of light for routine activities in the tent. a. Find an expression for the induced current as a function of time if you turn the crank at frequency \(f\). Assume that the semicircle is at its highest point at \(t=0\) s. b. With what frequency will you have to turn the crank for the maximum current to fully light the bulb? Is this feasible?

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