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Chapter 29: Problem 6

A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B =\) (0.0120 T/s)\(t\) + (3.00 \(\times\) 10\(^{-5}\) T/s\(^4)t^4\). 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 s?

Short Answer

Expert verified
(a) \(\varepsilon(t) = -2.512(0.0120 + 1.2 \times 10^{-4} t^3)\). (b) \(I = -1.131 \times 10^{-4}\) A at \(t = 5.00 \, \text{s}\).

Step by step solution

01

Identify the Variables

We are given a coil with a radius of 4.00 cm (converted to 0.04 m for calculations), 500 turns, and a magnetic field expressed as \(B(t) = (0.0120 \text{ T/s})t + (3.00 \times 10^{-5} \text{ T/s}^4)t^4\). The coil is also connected to a 600-\(\Omega\) resistor. We need to find the induced emf in the coil as a function of time and the current at \(t = 5.00 \text{ s}\).
02

Calculate the Area of the Coil

The area \(A\) of the coil can be calculated using the formula for the area of a circle: \(A = \pi r^2\). With \(r = 0.04 \text{ m}\), the area is:\[A = \pi (0.04)^2 = 5.024 \times 10^{-3} \text{ m}^2.\]
03

Determine the Induced EMF Using Faraday's Law

The induced emf \(\varepsilon\) in a coil is given by Faraday's Law: \(\varepsilon = -N \frac{d\Phi}{dt}\), where \(N\) is the number of turns, \(\Phi\) is the magnetic flux \(\Phi = B(t) \cdot A\). Thus:\[\varepsilon(t) = -N \frac{d}{dt}(B(t) \cdot A)\]\[\varepsilon(t) = -500 \times \frac{d}{dt}\left((0.0120t + 3.00 \times 10^{-5} t^4) \cdot 5.024 \times 10^{-3}\right).\]
04

Differentiate the Magnetic Flux

To find the rate of change of magnetic flux \(\Phi(t)\), perform the differentiation with respect to \(t\):\[\frac{d\Phi}{dt} = \frac{d}{dt}\left((0.0120t + 3.00 \times 10^{-5} t^4) \cdot 5.024 \times 10^{-3} \right).\]Simplifying,\[\frac{d\Phi}{dt} = 5.024 \times 10^{-3} \left(0.0120 + 4 \times 3.00 \times 10^{-5} t^3\right) = 5.024 \times 10^{-3} (0.0120 + 1.2 \times 10^{-4} t^3).\]
05

Calculate the Induced EMF Function

Substitute \(\frac{d\Phi}{dt}\) into the expression for the induced emf, resulting in:\[\varepsilon(t) = -500 \times 5.024 \times 10^{-3} (0.0120 + 1.2 \times 10^{-4} t^3).\]Simplifying, we get:\[\varepsilon(t) = -2.512(0.0120 + 1.2 \times 10^{-4} t^3).\]
06

Calculate Current at \(t = 5.00 \, \text{s}\)

Using Ohm's Law, \(I = \frac{\varepsilon}{R}\), where \(R = 600\, \Omega\), first find \(\varepsilon(5)\):\[\varepsilon(5) = -2.512(0.0120 + 1.2 \times 10^{-4} \times 5^3).\]\(\varepsilon(5) = -2.512(0.0120 + 0.015) = -2.512 \times 0.027 = -0.06784 \, \text{V}.\)Hence,\[I = \frac{-0.06784}{600} = -1.131 \times 10^{-4} \, \text{A}.\]

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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 outlines how a changing magnetic field induces an electromotive force (emf) in a coil, which is fundamental to various applications like power generation and electric circuits. Imagine a coil placed in a magnetic field. If this field changes over time, it will create an emf within the coil. In formula form, Faraday's Law is expressed as:
  • \[ \varepsilon = -N \frac{d\Phi}{dt} \]
Here, \(\varepsilon\) represents the induced emf, \(N\) is the number of turns in the coil, and \(\frac{d\Phi}{dt}\) is the rate of change of magnetic flux through the coil.

This negative sign adheres to Lenz's Law, which states that the induced emf will generate a current that opposes the change in magnetic flux.

In the given exercise, by plugging in the magnetic field's changing formula and performing the calculations, we derived \(\varepsilon(t) = -2.512(0.0120 + 1.2 \times 10^{-4} t^3)\), demonstrating how the change in the magnetic field over time can influence the induced emf in the coil.
Magnetic Flux
Magnetic flux measures the quantity of the magnetic field passing through a given area, with its comprehension being crucial in understanding electromagnetic induction. The magnetic flux \(\Phi\) for a flat circle in a uniform magnetic field is defined by:
  • \[ \Phi = B \cdot A \]
Where \(B\) is the magnetic field strength, and \(A\) is the area through which the magnetic field lines pass.

For a coil perpendicular to a magnetic field, as in our exercise, the flux is simplified and doesn't include any angle component because the angle between the magnetic field lines and the normal to the surface is zero.

In the exercise, with a magnetic field \(B(t)\) as a function of time and an area \(A\) derived using the formula for a circle, the magnetic flux \(\Phi(t) = B(t) \times A\). Differentiating \(\Phi\) with respect to time gives us the rate of change crucial for finding the induced emf.
Ohm's Law
Ohm's Law is a fundamental principle within electronics, relating voltage, current, and resistance within a circuit. It is essential when solving for current in circuits involving induced emfs. Ohm's Law is expressed as:
  • \[ I = \frac{V}{R} \]
Where \(I\) is the current, \(V\) is the voltage, and \(R\) is the resistance.

When an emf is induced in a coil connected to a resistor, as in the exercise, you can determine the resulting current passing through the resistor using this law. By substituting the calculated induced emf (from Faraday's Law) into the formula, you can derive the current.

In the exercise, at time \(t = 5.00\) seconds, the emf was found to be \(-0.06784\) V. By applying Ohm's Law with a resistor of 600 \(\Omega\), the current is calculated as \(-1.131 \times 10^{-4}\, \text{A}\), showcasing how Ohm鈥檚 Law is pivotal in determining current flow from induced emf.

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Most popular questions from this chapter

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\(^{-5}\) 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 V and estimate that you can rotate the coil at 30 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 long, straight solenoid with a cross-sectional area of 8.00 cm\(^2\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

A 25.0-cm-long metal rod lies in the \(xy\)-plane and makes an angle of 36.9\(^\circ\) with the positive \(x\)-axis and an angle of 53.1\(^\circ\) with the positive \(y\)-axis. The rod is moving in the \(+x\)-direction with a speed of 6.80 m/s. The rod is in a uniform magnetic field \(\overrightarrow{B} =\) (0.120 T)\(\hat{\imath}\) - (0.220 T)\(\hat{\jmath}\) - (0.0900 T)\(\hat{k}\). (a) What is the magnitude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

A circular conducting ring with radius \(r_0 =\) 0.0420 m lies in the xy-plane in a region of uniform magnetic field \(\overrightarrow{B} = B_0 [1 - 3(t/t_0)^2 + 2(t/t_0)^3]\hat{k}\). In this expression, \(t_0 =\) 0.0100 s and is constant, \(t\) is time, \(\hat{k}\) is the unit vector in the +\(z\)-direction, and \(B_0\) = 0.0800 T and is constant. At points \(a\) and \(b\) (Fig. P29.58) there is a small gap in the ring with wires leading to an external circuit of resistance \(R =\) 12.0 \(\Omega\). There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux \(\Phi_B\) through the ring. (b) Determine the emf induced in the ring at time \(t =\) 5.00 \(\times\) 10\(^{-3}\) s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through \(R\) at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time \(t =\) 1.21 \(\times\) 10\(^{-2}\) s. What is the polarity of the emf? (e) Determine the time at which the current through \(R\) reverses its direction.

A square, conducting, wire loop of side L, total mass m, and total resistance R initially lies in the horizontal xy-plane, with corners at (\(x, y, z\)) = (0, 0, 0), (0, \(L\), 0), (\(L\), 0, 0), and (\(L, L\), 0). There is a uniform, upward magnetic field \(\overrightarrow{B}\) = B\(\hat{k}\) in the space within and around the loop. The side of the loop that extends from (0, 0, 0) to (\(L\), 0, 0) is held in place on the \(x\)-axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the \(net\) torque (magnitude and direction) that acts on the loop when it has rotated through an angle \(\phi\) from its original orientation and is rotating downward at an angular speed \(\omega\). (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through 90\(^\circ\) ? Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.
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