13.2 Lenz's law (Page 2/3)

University physics volume 2 Page 2 / 3

Figure A shows a magnet that is moving towards the loop with the North pole facing the loop. The magnetic field lines leave the North pole of the magnet and cause the counterclockwise current flow in the loop. Figure B shows a magnet that is moving towards the loop with the South pole facing the loop. The magnetic field lines enter the South pole of the magnet and cause the clockwise current flow in the loop. — The change in magnetic flux caused by the approaching magnet induces a current in the loop. (a) An approaching north pole induces a counterclockwise current with respect to the bar magnet. (b) An approaching south pole induces a clockwise current with respect to the bar magnet.

Part (b) of the figure shows the south pole of a magnet moving toward a conducting loop. In this case, the flux through the loop due to the field of the magnet increases because the number of field lines directed from the back to the front of the loop is increasing. To oppose this change, a current is induced in the loop whose field lines through the loop are directed from the front to the back. Equivalently, we can say that the current flows in a direction so that the face of the loop nearer the approaching magnet is a south pole, which then repels the approaching south pole of the magnet. By RHR-2, your thumb points away from the bar magnet. Your fingers wrap in a clockwise fashion, which is the direction of the induced current.

Another example illustrating the use of Lenz’s law is shown in [link] . When the switch is opened, the decrease in current through the solenoid causes a decrease in magnetic flux through its coils, which induces an emf in the solenoid. This emf must oppose the change (the termination of the current) causing it. Consequently, the induced emf has the polarity shown and drives in the direction of the original current. This may generate an arc across the terminals of the switch as it is opened.

Figure A shows a circuit that consists of a solenoid, capacitor, and a closed switch. There is no current flow in the circuit. Figure B shows a circuit that consists of a solenoid, capacitor, and an opening switch. There is a current flow in the circuit. Figure C is a photo of an electrical arc generated between two metal contacts. — (a) A solenoid connected to a source of emf. (b) Opening switch S terminates the current, which in turn induces an emf in the solenoid. (c) A potential difference between the ends of the sharply pointed rods is produced by inducing an emf in a coil. This potential difference is large enough to produce an arc between the sharp points.

Check Your Understanding Find the direction of the induced current in the wire loop shown below as the magnet enters, passes through, and leaves the loop.

To the observer shown, the current flows clockwise as the magnet approaches, decreases to zero when the magnet is centered in the plane of the coil, and then flows counterclockwise as the magnet leaves the coil.
Figure shows a magnet that is moving into and through the loop with the South pole facing the loop. Position (a) corresponds to magnet approaching the loop; position (b) corresponds to the magnet directly into the loop. Position (c) corresponds to the magnet moving away from the loop.

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Check Your Understanding Verify the directions of the induced currents in [link] .

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A circular coil in a changing magnetic field

A magnetic field

\vec{B}

is directed outward perpendicular to the plane of a circular coil of radius

r = 0.50 m

( [link] ). The field is cylindrically symmetrical with respect to the center of the coil, and its magnitude decays exponentially according to

B = (1.5 T) e^{− (5.0 s^{−1}) t},

where B is in teslas and t is in seconds. (a) Calculate the emf induced in the coil at the times

t_{1} = 0,

t_{2} = 5.0 \times 1 0^{− 2} s,

and

t_{3} = 1.0 s .

(b) Determine the current in the coil at these three times if its resistance is

10 Ω .

Figure shows a circular coil of radius r in a decreasing uniform magnetic field. — A circular coil in a decreasing magnetic field.

Strategy

Since the magnetic field is perpendicular to the plane of the coil and constant over each spot in the coil, the dot product of the magnetic field

\vec{B}

and normal to the area unit vector

\hat{n}

turns into a multiplication. The magnetic field can be pulled out of the integration, leaving the flux as the product of the magnetic field times area. We need to take the time derivative of the exponential function to calculate the emf using Faraday’s law. Then we use Ohm’s law to calculate the current.

Solution

Since $\vec{B}$ is perpendicular to the plane of the coil, the magnetic flux is given by

$\begin{matrix} Φ_{m} & = B π r^{2} = (1.5 e^{−5.0 t} T) π {(0.50 m)}^{2} \\ = 1.2 e^{− (5.0 s^{−1}) t} Wb . \end{matrix}$

From Faraday’s law, the magnitude of the induced emf is

$ε = | \frac{d Φ_{m}}{d t} | = | \frac{d}{d t} (1.2 e^{− (5.0 s^{−1}) t} Wb) | = 6.0 e^{− (5.0 s^{−1}) t} V .$

Since $\vec{B}$ is directed out of the page and is decreasing, the induced current must flow counterclockwise when viewed from above so that the magnetic field it produces through the coil also points out of the page. For all three times, the sense of ε is counterclockwise; its magnitudes are

$ε (t_{1}) = 6.0 V; ε (t_{2}) = 4.7 V; ε (t_{3}) = 0.040 V .$
From Ohm’s law, the respective currents are

$\begin{matrix} I (t_{1}) & = & \frac{ε (t_{1})}{R} = \frac{6.0 V}{10 Ω} = 0.60 A; \\ I (t_{2}) & = & \frac{4.7 V}{10 Ω} = 0.47 A; \end{matrix}$

and

$I (t_{3}) = \frac{0.040 V}{10 Ω} = 4.0 \times 10^{−3} A .$

Significance

An emf voltage is created by a changing magnetic flux over time. If we know how the magnetic field varies with time over a constant area, we can take its time derivative to calculate the induced emf.

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Changing magnetic field inside a solenoid

The current through the windings of a solenoid with

n = 2000

turns per meter is changing at a rate

d I / d t = 3.0 A / s .

(See Sources of Magnetic Fields for a discussion of solenoids.) The solenoid is 50-cm long and has a cross-sectional diameter of 3.0 cm. A small coil consisting of

N = 20

closely wound turns wrapped in a circle of diameter 1.0 cm is placed in the middle of the solenoid such that the plane of the coil is perpendicular to the central axis of the solenoid. Assuming that the infinite-solenoid approximation is valid at the location of the small coil, determine the magnitude of the emf induced in the coil.

Strategy

The magnetic field in the middle of the solenoid is a uniform value of

μ_{0} n I .

This field is producing a maximum magnetic flux through the coil as it is directed along the length of the solenoid. Therefore, the magnetic flux through the coil is the product of the solenoid’s magnetic field times the area of the coil. Faraday’s law involves a time derivative of the magnetic flux. The only quantity varying in time is the current, the rest can be pulled out of the time derivative. Lastly, we include the number of turns in the coil to determine the induced emf in the coil.

Solution

Since the field of the solenoid is given by

B = μ_{0} n I,

the flux through each turn of the small coil is

Φ_{m} = μ_{0} n I (\frac{π d^{2}}{4}),

where d is the diameter of the coil. Now from Faraday’s law, the magnitude of the emf induced in the coil is

\begin{matrix} ε & = | N \frac{d Φ_{m}}{d t} | = | N μ_{0} n \frac{π d^{2}}{4} \frac{d I}{d t} | \\ = 20 (4 π \times 10^{−7} T \cdot m/s) (2000 m^{-1}) \frac{π {(0.010 m)}^{2}}{4} (3.0 A/s) \\ = 1.2 \times 10^{−5} V . \end{matrix}

Significance

When the current is turned on in a vertical solenoid, as shown in [link] , the ring has an induced emf from the solenoid’s changing magnetic flux that opposes the change. The result is that the ring is fired vertically into the air.

Figure is a photograph of a thin metal ring levitating above the vertical solenoid. — The jumping ring. When a current is turned on in the vertical solenoid, a current is induced in the metal ring. The stray field produced by the solenoid causes the ring to jump off the solenoid.

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Visit this website for a demonstration of the jumping ring from MIT.

Summary

We can use Lenz’s law to determine the directions of induced magnetic fields, currents, and emfs.
The direction of an induced emf always opposes the change in magnetic flux that causes the emf, a result known as Lenz’s law.

Conceptual questions

The circular conducting loops shown in the accompanying figure are parallel, perpendicular to the plane of the page, and coaxial. (a) When the switch S is closed, what is the direction of the current induced in D ? (b) When the switch is opened, what is the direction of the current induced in loop D ?

Figure shows a circuit that consists of a resistor, capacitor, opened switch and a loop C. A loop D is located next to the loop C.

a. CW as viewed from the circuit; b. CCW as viewed from the circuit

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The north pole of a magnet is moved toward a copper loop, as shown below. If you are looking at the loop from above the magnet, will you say the induced current is circulating clockwise or counterclockwise?

Figure shows a magnet that is moving towards the loop with the North pole facing the loop.

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The accompanying figure shows a conducting ring at various positions as it moves through a magnetic field. What is the sense of the induced emf for each of those positions?

Figure shows the loop that moves through the area with the magnetic field perpendicular to it.

As the loop enters, the induced emf creates a CCW current while as the loop leaves the induced emf creates a CW current. While the loop is fully inside the magnetic field, there is no flux change and therefore no induced current.

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Show that $ε$ and $d Φ_{m} / d t$ have the same units.

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State the direction of the induced current for each case shown below, observing from the side of the magnet.

Figure A shows a magnet that is moving towards the loop with the North pole facing the loop. Figure B shows a magnet that is moving away from the loop with the North pole facing the loop. Figure C shows a magnet that is moving towards the loop with the South pole facing the loop. Figure D shows a magnet that is hold constant with the North pole facing the loop. The loop is moving towards the magnet. Figure E shows a magnet that is hold constant with the North pole facing the loop. Loop is moving away from the magnet. Figure F shows a magnet that is hold constant with the North pole facing the loop. Loop is rotating clockwise.

a. CCW viewed from the magnet; b. CW viewed from the magnet; c. CW viewed from the magnet; d. CCW viewed from the magnet; e. CW viewed from the magnet; f. no current

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Problems

A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10-s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. (a) Determine the emf induced in the loop. (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop?

a. $7.8 \times 10^{−3} V$ ; b. CCW from the same view as the magnetic field

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When a magnetic field is first turned on, the flux through a 20-turn loop varies with time according to $Φ_{m} = 5.0 t^{2} - 2.0 t,$ where $Φ_{m}$ is in milliwebers, t is in seconds, and the loop is in the plane of the page with the unit normal pointing outward. (a) What is the emf induced in the loop as a function of time? What is the direction of the induced current at (b) t = 0, (c) 0.10, (d) 1.0, and (e) 2.0 s?

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The magnetic flux through the loop shown in the accompanying figure varies with time according to $Φ_{m} = 2.00 e^{−3 t} sin (120 π t),$ where $Φ_{m}$ is in milliwebers. What are the direction and magnitude of the current through the $5.00 - Ω$ resistor at (a) $t = 0$ ; (b) $t = 2.17 \times 1 0^{− 2} s,$ and (c) $t = 3.00 s?$

Figure shows a loop with the magnetic flux perpendicular to the loop. Loop is connected to a 5 Ohm resistor.

a. 150 A downward through the resistor; b. 232 A upward through the resistor; c. 0.093 A downward through the resistor

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Use Lenz’s law to determine the direction of induced current in each case.

Figure A shows a metal bar moving to the left in the perpendicular uniform magnetic field. Figure B shows a loop moving to the right in a parallel uniform magnetic field. Figure C shows a loop moved into a perpendicular uniform magnetic field. Figure D shows a metal bar moving to the right in the perpendicular uniform magnetic field. Figure E shows a loop located into an increasing perpendicular magnetic field. Figure F shows a loop located into a decreasing parallel magnetic field.

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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