30.5.1 Using Faraday’s Law

Most electromagnetic induction problems can be solved with a four-step strategy.

PROBLEM-SOLVING STRATEGY 30.1
Electromagnetic induction

MODEL Make simplifying assumptions about wires and magnetic fields.

VISUALIZE Draw a picture or a circuit diagram. Use Lenz’s law to determine the direction of the induced current.

SOLVE The mathematical representation is based on Faraday’s law

ε = | \frac{d Φ_{m}}{d t} |

For an N-turn coil, multiply by N. The size of the induced current is $I = ε / R$ .

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercise 18

EXAMPLE 30.8 Electromagnetic induction in a solenoid

A 2.0-cm-diameter loop of wire with a resistance of $0.010 Ω$ is placed in the center of the solenoid seen in FIGURE 30.27a . The solenoid is 4.0 cm in diameter, 20 cm long, and wrapped with 1000 turns of wire. FIGURE 30.27b shows the current through the solenoid as a function of time as the solenoid is “powered up.” A positive current is defined to be cw when seen from the left. Find the current in the loop as a function of time and show the result as a graph.

The figure illustrates a loop inside the envelope of a solenoid.

The figure illustrates the traversal of a linear curve, on a coordinate plane. The horizontal axis is labeled t (seconds), and the vertical axis is labeled I subscript sol (Ampere).

Full Alternative Text

MODEL The solenoid’s length is much greater than its diameter, so the field near the center should be nearly uniform.

VISUALIZE The magnetic field of the solenoid creates a magnetic flux through the loop of wire. The solenoid current is always positive, meaning that it is cw as seen from the left. Consequently, from the right-hand rule, the magnetic field inside the solenoid always points to the right. During the first second, while the solenoid current is increasing, the flux through the loop is to the right and increasing. To oppose the change in the flux, the loop’s induced magnetic field must point to the left. Thus, again using the right-hand rule, the induced current must flow ccw as seen from the left. This is a negative current. There’s no change in the flux for $t > 1 s$ , so the induced current is zero.

SOLVE Now we’re ready to use Faraday’s law to find the magnitude of the current. Because the field is uniform inside the solenoid and perpendicular to the loop $(θ = 0 °)$ , the flux is $Φ_{m} = A B$ , where $A = π r^{2} = 3.14 \times 10^{- 4} m^{2}$ is the area of the loop (not the area of the solenoid). The field of a long solenoid of length l was found in Chapter 29 to be

B = \frac{μ_{0} N I_{sol}}{l}

The flux when the solenoid current is $I_{sol}$ is thus

Φ_{m} = \frac{μ_{0} A N I_{sol}}{l}

The changing flux creates an induced emf $ε$ that is given by Faraday’s law:

ε = | \frac{d Φ_{m}}{d t} | = \frac{μ_{0} A N}{l} | \frac{d I_{sol}}{d t} | = 2.0 \times 10^{- 6} | \frac{d I_{sol}}{d t} |

From the slope of the graph, we find

| \frac{d I_{sol}}{d t} | = {\begin{matrix} 10 A / s & 0.0 s < t < 1.0 s \\ 0 A / s & 1.0 s < t < 3.0 s \end{matrix}

Thus the induced emf is

ε = {\begin{matrix} 2.0 \times 10^{- 5} V & 0.0 s < t < 1.0 s \\ 0 V & 1.0 s < t < 3.0 s \end{matrix}

Finally, the current induced in the loop is

I_{loop} = \frac{ε}{R} = {\begin{matrix} - 2.0 mA & 0.0 s < t < 1.0 s \\ 0 mA & 1.0 s < t < 3.0 s \end{matrix}

where the negative sign comes from Lenz’s law. This result is shown in FIGURE 30.28.

FIGURE 30.28 Full Alternative Text

EXAMPLE 30.9 Current induced by an MRI machine

The body is a conductor, so rapid magnetic field changes in an MRI machine can induce currents in the body. To estimate the size of these currents, and any biological hazard they might impose, consider the “loop” of muscle tissue shown in FIGURE 30.29. This might be muscle circling the bone of your arm or thigh. Although muscle is not a great conductor—its resistivity is $1.5 Ω m$ —we can consider it to be a conducting loop with a rather high resistance. Suppose the magnetic field along the axis of the loop drops from 1.6 T to 0 T in 0.30 s, which is about the largest possible rate of change for an MRI solenoid. What current will be induced?

The figure illustrates the edge view of a circular loop in a magnetic field.

FIGURE 30.29 Full Alternative Text

MODEL Model the muscle as a conducting loop. Assume that $B$ decreases linearly with time.

SOLVE The magnetic field is parallel to the axis of the loop, with $θ = 0 °$ , so the magnetic flux through the loop is $Φ_{m} = A B = π r^{2} B$ . The flux changes with time because $B$ changes. According to Faraday’s law, the magnitude of the induced emf is

ε = | \frac{d Φ_{m}}{d t} | = π r^{2} | \frac{d B}{d t} |

The rate at which the magnetic field changes is

\frac{d B}{d t} = \frac{Δ B}{Δ t} = \frac{- 1.60 T}{0.30 s} = - 5.3 T / s

$d B / d t$ is negative because the field is decreasing, but all we need for Faraday’s law is the absolute value. Thus

ε = π r^{2} | \frac{d B}{d t} | = π (0.040 {m)}^{2} (5.3 T / s) = 0.027 V

To find the current, we need to know the resistance of the loop. Recall, from Chapter 27, that a conductor with resistivity $ρ$ , length $L$ , and cross-section area $A$ has resistance $R = ρ L / A$ . The length is the circumference of the loop, calculated to be $L = 0.25 m,$ and we can use the 1.0 cm diameter of the “wire” to find $A = 7.9 \times 10^{- 5} m^{2}$ . With these values, we can compute $R = 4700 Ω$ . As a result, the induced current is

I = \frac{ε}{R} = \frac{0.027 V}{4700 Ω} = 5.7 \times 10^{- 6} A = 5.7 μ A

REVIEW This is a very small current. Power—the rate of energy dissipation in the muscle—is

P = I^{2} R = (5.7 \times 10^{- 6} {A)}^{2} (4700 Ω) = 1.5 \times 10^{- 7} W

The current is far too small to notice, and the tiny energy dissipation will certainly not heat the tissue.

NOTE We sometimes use R to represent the radius of a loop or coil. Be careful not to confuse a dimension R with resistance R.

30.5.1 Using Faraday’s Law

PROBLEM-SOLVING STRATEGY 30.1
Electromagnetic induction

EXAMPLE 30.8 Electromagnetic induction in a solenoid

FIGURE 30.27 A loop inside a solenoid.

FIGURE 30.28 The induced current in the loop.

EXAMPLE 30.9 Current induced by an MRI machine

FIGURE 30.29 Edge view of a loop of muscle tissue in a magnetic field.