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H10 Chapter 29: Motional EMF & Maxwell's EquationsDue: 11:59pm on Monday, November 11, 2013

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Faraday's Law and Induced EmfDescription: Discusses Faraday's law; presents a sequence of questions related to finding the induced emf under different circumstances.

Learning Goal:

To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf.

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as

,where is the emf induced in a closed loop, and

is the rate of change of the magnetic flux through a surface bounded by the loop. For uniform magnetic fields the magnetic flux is given by

, where is the angle between the magnetic field and the normal to the surface of area .

To find the direction of the induced emf, one can use Lenz's law:

The induced current's magnetic field opposes the change in the magnetic flux that induced the current.

For example, if the magnetic flux through a loop increases, the induced magnetic field is directed opposite to the "parent" magnetic field, thus countering theincrease in flux. If the flux decreases, the induced current's magnetic field has the same direction as the parent magnetic field, thus countering the decreasein flux.

Recall that to relate the direction of the electric current and its magnetic field, you can use the right-hand rule: When the fingers on your right hand arecurled in the direction of the current in a loop, your thumb gives the direction of the magnetic field generated by this current.

In this problem, we will consider a rectangular loop of wire with sides and placed in a region where a uniform magnetic field exists (see the diagram).The resistance of the loop is .

Initially, the field is perpendicular to the plane of the loop and is directed out of the page. The loopcan rotate about either the vertical or horizontal axis, passing through the midpoints of theopposite sides, as shown.

Part A

Which of the following changes would induce an electromotive force (emf) in the loop? When you consider each option, assume that no other changesoccur.

Check all that apply.

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H10 Chapter 29: Motional EMF & Maxwell's Equations [ Edit ]

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E = − ∆ΦB

∆tE

∆ΦB

∆t

= ⋅ = BAcos(θ)ΦB B A θ B A

x y B R

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ANSWER:

Part B

Find the flux through the loop.

Express your answer in terms of , , and .

ANSWER:

Part C

If the magnetic field steadily decreases from to zero during a time interval , what is the magnitude of the induced emf?

Express your answer in terms of , , , and .

Hint 1. Find the change in magnetic flux

What is the total change in magnetic flux during this time interval?

Express your answer in terms of , , and .

ANSWER:

ANSWER:

Part D

If the magnetic field steadily decreases from to zero during a time interval , what is the magnitude of the induced current?

Express your answer in terms of , , , , and the resistance of the wire.

ANSWER:

Part E

If the magnetic field steadily decreases from to zero during a time interval , what is the direction of the induced current?

ANSWER:

The magnitude of increases.

The magnitude of decreases.

The loop rotates about the vertical axis (vertical dotted line) shown in the diagram.

The loop rotates about the horizontal axis (horizontal dotted line) shown in the diagram.

The loop moves to the right while remaining in the plane of the page.

The loop moves toward you, out of the page, while remaining parallel to itself.

B

B

ΦB

x y B

= ΦB

B t E

x y B t

∆ΦB

x y B

= ∆ΦB

= E

B t I

x y B t R

= I

B t

The flux decreases, so the induced magnetic field must be in the same direction as the original (parent) magnetic field. Therefore, the inducedmagnetic field is out of the page. Using the right-hand rule, we deduce that the direction of the current is counterclockwise.

Part F

Which of the following changes would result in a clockwise emf in the loop? When you consider each option, assume that no other changes occur.

Check all that apply.

ANSWER:

Clockwise emf implies that the induced magnetic field is directed into the page. Therefore, the magnetic flux of the original field must beincreasing. Only the first option corresponds to increasing flux.

Induced EMF and Current in a Shrinking LoopDescription: Find the induced emf and the direction of the induced current in a loop whose circumference is decreasing with time.

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 166 , but its circumference is decreasing at a constant rate of 14.0 due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 1.00 , which is oriented perpendicular to the

plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.

Part A

Find the magnitude of the emf induced in the loop after exactly time 9.00 has passed since the circumference of the loop started to decrease.

Express your answer numerically in volts to three significant figures.

Hint 1. How to approach the problem

The induced emf in a loop is related to the rate of change of magnetic flux through the loop. (If you don't know the exact relation, look at HintA.3.) Therefore, to find the induced emf, you first need to find the magnetic flux through the loop as a function of time. The magnetic flux throughthe loop is proportional to the area of the loop, so a good starting point is to write the area of the loop as a function of time in terms of the givenparameters. The area of the loop is related to its radius, which in turn can be related to its circumference.

Hint 2. An expression for the circumference of the loop as a function of time

Let be the initial circumference of the coil. At time , the circumference starts decreasing at the constant rate . Then the circumferenceof the coil as a function of time is given by the relation .

Hint 3. An expression for the flux through the loop as a function of its circumference

The equation for the magnetic flux through the loop is

,

where is the magnetic field through the loop (which is a constant in this problem), is the area vector associated with the loop, and isthe angle between the magnetic field vector and the area vector.

In this problem, is given as (the area vector is parallel to the magnetic field vector). Therefore,

clockwise

counterclockwise

The magnitude of increases.

The magnitude of decreases.

The loop rotates through 45 degrees about the vertical axis (vertical dotted line) shown in the diagram.

The loop rotates through 45 degrees about the horizontal axis (horizontal dotted line) shown in the diagram.

The loop moves to the right while remaining in the plane of the page.

The loop moves toward you, out of the page, while remaining parallel to itself.

B

B

cmcm/s T

E s

C0 t = 0 a

t C(t) = − atC0

Φ

Φ(t) = ⋅ (t) = (t) cos(ϕ)B A ∣∣B ∣∣∣∣A

∣∣

B (t)A ϕ

ϕ 0∘

Φ(t) = (t) = Bπ[r(t)∣∣ ∣∣∣∣

∣∣

2

,

where and is the radius of the loop as a function of time.

Finally, the radius of the loop is related to its circumference by . Substituting this relation into the equation for gives

.

Hint 4. A formula for the induced emf in the loop (Faraday's law)

The emf induced in the loop (Faraday's law) is

,

where is the magnetic flux through the loop.

Hint 5. An expression for

The chain rule of calculus allows us to express , where is a function of , as follows:

.

ANSWER:

Part B

Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

ANSWER:

The induced current flows in the direction that tends to prevent the flux through the coil from decreasing. That is, it adds to the magnetic fieldthrough the coil as the coil's area is decreasing. This means that the current has to flow clockwise, so that the magnetic field produced by it (right-hand rule) points away from you (you were asked to look at the loop along the direction of the original magnetic field). Alternatively, you could lookat how each part of the wire moves toward the center of the loop as it gets smaller. As a result, we can use the standard equation for force on aparticle and the right-hand rule to determine the direction of the current.

Exercise 29.4Description: A closely wound search coil has an area of A, N turns, and a resistance of R_1. It is connected to a charge-measuring instrument whoseresistance is R_2. When the coil is rotated quickly from a position parallel to a uniform magnetic field to one...

A closely wound search coil has an area of 3.18 , 130 turns, and a resistance of 60.7 . It is connected to a charge-measuring instrument whoseresistance is 45.9 . When the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrumentindicates a charge of 3.56×10−5 .

Part A

What is the magnitude of the field?

ANSWER:

Φ(t) = (t) = Bπ[r(t)∣∣B ∣∣∣∣A

∣∣ ]2

B = | |B r(t)

r(t) C(t) r(t) = C(t)2π

Φ

Φ(t) = B[C(t)]2

4π

E

E = − dΦdt

Φ

dx2

dt

dx2

dtx t

= 2xdx2

dtdxdt

= = 8.91×10−3 V E

clockwise

counterclockwise

F = q ×v B

cm2 ΩΩ

C

Exercise 29.7Description: The current in the long, straight wire AB shown in the figure is upward and is increasing steadily at a rate (di)/(dt). (a) At an instant whenthe current is i, what is the magnitude of the field B_vec at a distance r to the right of the wire? (b)...

The current in the long, straight wire AB shown in the figure is upward and is increasing steadilyat a rate .

Part A

At an instant when the current is , what is the magnitude of the field at a distance to the right of the wire?

Express your answer in terms of the appropriate quantities.

ANSWER:

Part B

At an instant when the current is , what is the direction of the field at a distance to the right of the wire?

ANSWER:

Part C

What is the flux through the narrow shaded strip?

Express your answer in terms of the appropriate quantities.

ANSWER:

Part D

What is the total flux through the loop?

Express your answer in terms of the appropriate quantities.

= = 9.18×10−2 B T

didt

i B r

= B

i B r

into the page

out of the page

dΦB

= dΦB

ANSWER:

Part E

What is the induced emf in the loop?

Express your answer in terms of the appropriate quantities.

ANSWER:

Part F

Evaluate the numerical value of the induced emf if , , , and .

ANSWER:

Problem 29.14Description: The armature of a small generator consists of a flat, square coil with sides of length L and with N turns. The coil rotates in a magnetic fieldof magnitude B. (a) What is the angular speed of the coil if the maximum emf produced is EMF?

The armature of a small generator consists of a flat, square coil with sides of length 1.05 and with 100 turns. The coil rotates in a magnetic field ofmagnitude 0.795 .

Part A

What is the angular speed of the coil if the maximum emf produced is 2.80×10−3 ?

ANSWER:

Problem 29.20Description: A long metal bar of length L is pulled to the right at a steady speed v perpendicular to a uniform magnetic field of magnitude B. The barrides on parallel metal rails connected through a resistance of R, as shown in the figure , so the apparatus...

A long metal bar of length 1.90 is pulled to the right at a steady speed 5.20 perpendicular to a uniform magnetic field of magnitude = 0.760 .The bar rides on parallel metal rails connected through a resistance of = 25.3 , as shown in the figure , so the apparatus makes a complete circuit. Theresistance of the bar and the rails can be neglected.

= Φ

= E

a = 12.0 cm b = 36.0 cm L = 24.0 cm di/dt = 9.60A/s

= 5.06×10−7 E V

cmT

V

= 0.319 rad/s

m m/s B TR Ω

Part A

Calculate the magnitude of the emf induced in the circuit.

ANSWER:

Part B

Find the direction of the current induced in the circuit.

ANSWER:

Part C

Calculate the current through the resistor.

ANSWER:

Problem 29.30Description: A long, thin solenoid has N turns per meter and a radius of r_1. The current in the solenoid is increasing at a uniform rate di/dt. Themagnitude of the induced electric field at a point which is near the center of the solenoid and a distance of r_2...

A long, thin solenoid has 440 turns per meter and a radius of 1.00 . The current in the solenoid is increasing at a uniform rate di/dt. The magnitude of theinduced electric field at a point which is near the center of the solenoid and a distance of 3.56 from its axis is 8.00×10−6 .

Part A

Calculate di/dt.

Use 1.26×10−6 for the permeability of free space.

ANSWER:

= 7.51 V

clockwise

counterclockwise

= 0.297 A

cmcm V/m

N/A2

= 10.3 A/s

Electric Field Due to Increasing FluxDescription: Step by step application of Faraday's law for B increasing linearly with time.

Learning Goal:

To work through a straightforward application of Faraday's law to find the EMF and the electric field surrounding a region of increasing flux

Faraday's law describes how electric fields and electromotive forces are generated from changing magnetic fields. This problem is a prototypical example inwhich an increasing magnetic flux generates a finite line integral of the electric field around a closed loop that surrounds the changing magnetic flux througha surface bounded by that loop. A cylindrical iron rod with cross-sectional area is oriented with its symmetry axis coincident with the z axis of a cylindricalcoordinate system as shown. It has a uniform magnetic field inside that varies according to . In other words, the magentic field isalways in the positive z direction, and it has no other components.

For your convenience, we restate Faraday's law here: ,

where is the line integral of the electric field, and the magnetic flux is given by

, where is the angle between the magnetic field andthe local normal to the surface bounded by the closed loop.

Direction: The line integral and surface integral reverse their signs if the reference direction of or is reversed. The right-hand rule applies here: If the thumb of your right hand is taken along

, then the fingers point along d\vecl. You are free to take the loop anywhere you choose,although usually it makes sense to choose it to lie along the path of the circuit you areconsidering.

Part A

Find \calE(R,t), the electromotive force (EMF) around a loop that is at distance \texttipRR from the z axis, where \texttipRR is restricted to theregion outside the iron rod as shown. Take the direction shown in the figure as positive.

Express \calE(R,t) in terms of \texttipAA, \texttipB_\rm 0B_0, \texttipB_\rm 1B_1, \texttipRR, and any needed constantssuch as \texttip\epsilon _\rm 0epsilon_0, \texttip\pi pi, and \texttip\mu _\rm 0mu_0.

Hint 1. Selecting the loop

Since you want to find E_\theta(R,t), but Faraday's law determines only an integral of the field, you must select the loop in such a way that itinvolves only one value of \texttipRR and has a constant projection of \vecE along it. This would be a circle with fixed \texttipRR (at any \texttipzz: the rod extends infinitely in the z direction).

Hint 2. Find the magnetic flux

Find the magnetic flux \texttip\Phi_B(t)Phi_B(t) at time \texttiptt.

Express \texttip\Phi_B(t)Phi_B(t) in terms of \texttipB_\rm 0B_0, \texttipB_\rm 1B_1, and other given quantities.

Hint 1. Flux integral

The flux integral for a constant field is the value of the magnetic field component perpendicular to the surface selected times the area, B(t) A. The area is the region in which the field is present and also is inside the loop selected.

ANSWER:

ANSWER:

A(t) = + tBz B0 B1

E = − = − ∮ ( ,t) ⋅ ddΦB

dtddt

B r A

E = ∮ ( , t) ⋅ dE r l

= ∮ ( , t) ⋅ d = ∮ B|dA| cos(θ)ΦB B r A θ

dl

dA

dA

\Phi_B(t) =

\calE(R,t) =

Part B

Due to the cylindrical symmetry of this problem, the induced electric field \texttip\vecE\left(R,t\right)E_vec(R,t) can depend only on the distance \texttipRR from the z axis, where \texttipRR is restricted to the region outside the iron rod. Find this field.

Express \texttip\vecE\left(R,t\right)E_vec(R,t) in terms of quantities given in the introduction (and constants), using the unit vectors inthe cylindrical coordinate system, \texttip\hat\theta theta_unit, \texttip\hatrr_unit, and \texttip\hatkk_unit.

Hint 1. Calculate the line integral

The component \texttipE_\rm \theta E_theta of the electric field is the only component picked out by the dot product in the line integral fordeterming the EMF, \calE(R, t) . Since d\vecl is always parallel to \texttip\hat\theta theta_unit, all other components of the electric fieldare perpendicular to d\vecl. What is \calE(R, t)?

Express \calE(R, t) in terms of \texttipE_\rm \theta E_theta, quantities given in the introduction to this part, and familiarconstants.

ANSWER:

Equate this to the expression from the previous part to find \texttipE_\rm \theta E_theta in terms of the magnetic field etc.

Hint 2. The z and r components of the electric field

All the magnetic induction fields are in the \texttip\theta theta direction. Any line integral along r or z will enclose zero magnetic flux, so theinduced electric field along these directions will be zero. (If the rod were charged, there would be a radial field, but the problem states that it isnot.)

ANSWER:

Exercise 29.38Description: A metal ring d in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to themagnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled...

A metal ring 4.10\rm cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magneticfield. These magnets produce an initial uniform field of 1.12 \rm T between them but are gradually pulled apart, causing this field to remain uniform butdecrease steadily at 0.200\rm T/s .

Part A

What is the magnitude of the electric field induced in the ring?

ANSWER:

Part B

In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

ANSWER:

\calE(R, t) =

\texttip\vecE\left(R,t\right)E_vec(R,t) =

E = = 2.05×10−3 \rm V/m

Problem 29.50Description: Consider the loop in the figure . The area is A, and it spins with angular velocity omega in a magnetic field of strength B. (a) What is themaximum induced emf if the loop is rotated about the y-axis? (b) What is the maximum induced emf if the...

Consider the loop in the figure . The area is \texttipAA = 630\rm cm^2 , and it spins withangular velocity \texttip\omega omega = 33.0\rm rad/s in a magnetic field of strength\texttipBB = 0.490\rm T .

Part A

What is the maximum induced emf if the loop is rotated about the y-axis?

ANSWER:

Part B

What is the maximum induced emf if the loop is rotated about the x-axis?

ANSWER:

Part C

What is the maximum induced emf if the loop is rotated about an edge parallel to the z-axis?

ANSWER:

Problem 29.63Description: A slender rod with a length of L rotates with an angular speed of omega about an axis through one end and perpendicular to the rod. Theplane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of B. (a) What is the...

A slender rod with a length of 0.200\rm m rotates with an angular speed of 8.10\rm rad/s about an axis through one end and perpendicular to the rod.The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.650\rm T .

Part A

What is the induced emf in the rod?

Counterclockwise

Clockwise

= 1.02 V

0 V

= 1.02 V

ANSWER:

Part B

What is the potential difference between its ends?

ANSWER:

Part C

Suppose instead the rod rotates at 8.10\rm rad/s about an axis through its center and perpendicular to the rod. In this case, what is the potentialdifference between the ends of the rod?

ANSWER:

Part D

Suppose instead the rod rotates at 8.10\rm rad/s about an axis through its center and perpendicular to the rod. In this case, what is the potentialdifference between the center of the rod and one end?

ANSWER:

± Motional EMF in a Conducting RodDescription: ± Includes Math Remediation. Finding the Electric field, potential difference, and the motional emf in a conducting rod moving in a uniformmagnetic field.

In the figure, a conducting rod with length \texttipLL = 35.0\rm cm moves in a magnetic field \texttip\vecB\vecB of magnitude 0.540\rm T directedinto the plane of the figure. The rod moves with speed \texttipvv = 7.00\rm m/s in the direction shown.

Part A

When the charges in the rod are in equilibrium, which point, a or b, has an excess of positive charge?

= 0.105 V

= 0.105 V

0 V

= 2.63×10−2 V

Hint 1. Finding the force on charges moving in a magnetic field

The force \texttip\vecFF_vec on a charge \texttipqq moving with velocity \texttip\vecvv_vec in a magnetic field \texttip\vecBB_vec is given by

\vecF=q \vecv \times \vecB.One can use the right-hand rule to determine the direction in which the charges in the rod are forced.

ANSWER:

Part B

In what direction does the electric field then point?

Hint 1. Where does the electric field come from?

Keep in mind that the electric field is due to the fact that the positive and negative charges have been separated due to the force from themagnetic field. Recall that electric fields start on positive charges and end on negative charges.

ANSWER:

Part C

When the charges in the rod are in equilibrium, what is the magnitude \texttipEE of the electric field within the rod?

Express your answer in volts per meter to at least three significant figures.

Hint 1. How to approach the problem

Use the fact that at equilibrium, the electric and magnetic forces on a test charge in the rod must be equal and opposite to each other.

Hint 2. Formula for the force on a charge moving in a magnetic field

The force \texttip\vecFF_vec on a charge \texttipqq moving with velocity \texttip\vecvv_vec in a magnetic field \texttip\vecBB_vec is given by

\vecF=q \vecv \times \vecB.

Hint 3. Formula for the force on a charge in an electric field

The force \texttip\vecFF_vec on a charge \texttipqq in an electric field \texttip\vecEE_vec is

\vecF = q \vecE.

Hint 4. Units

Recall that the SI unit of a magnetic field is the tesla, velocity is measured in meters per second, and electric field in volts per meter.

ANSWER:

a

b

from a toward b

from b toward a

\texttipEE = = 3.78 \rm V/m

Part D

Which point, a or b, is at higher potential?

Hint 1. Formula for the potential difference

The formula for the potential difference \texttipV_\rm baV_ba between two points a and b in a region of space containing an electric field \texttip\vecEE_vec is

\largeV_ba= -\int_b^a \vecE \cdot \vecdl,where the line integral can be taken along any path from b to a. In particular, the straight line connecting b to a (the ends of the rod), i.e., thepath along the length of the rod, is a good choice for this problem.

ANSWER:

Some work must have been done in order to create this potential difference, i.e., to separate the charges in the rod. This work was done by theinitial force required to pull the rod in opposition to the force on it due to the interaction of the transient current in it with the magnetic field. Notethat once the charges are in equilibrium, no force is required to keep the rod moving with constant velocity.

Part E

What is the magnitude \texttipV_\rm baV_ba of the potential difference between the ends of the rod?

Express your answer in volts to at least three significant figures.

Hint 1. Formula for the potential difference

The formula for the potential difference \texttipV_\rm baV_ba between two points a and b in a region of space containing an electric field \texttip\vecEE_vec is

\largeV_ba= -\int_b^a \vecE \cdot \vecdl,where the line integral can be taken along any path from b to a. In particular, the straight line connecting b to a (the ends of the rod), i.e., thepath along the length of the rod, is a good choice for this problem.

Hint 2. Calculating the integral

Note that the electric field \texttip\vecE\vecE in the rod is a constant (the rod is a conductor), and so the integral should be relativelystraightforward.

ANSWER:

Part F

What is the magnitude \cal E of the motional emf induced in the rod?

Express your answer in volts to at least three significant figures.

Hint 1. Definition of motional emf

Motional emf is the potential difference generated by moving a conducting rod through a magnetic field. The motional emf is measured betweenthe two ends of the rod.

ANSWER:

a

b

\texttipV_\rm baV_ba = = 1.32 \rm V

The Magnetic Field between Capacitor PlatesDescription: Find the magnetic field inside a charging capacitor using Ampère's law.

A capacitor consists of two parallel circular plates of radius \texttipLL. The capacitor has capacitance \texttipCC and is being charged in a simplecircuit loop. The circuit has an initial current \texttipI_\rm 0I_0 and consists of the capacitor, a battery with voltage \texttipVV, and a resistor withresistance \texttipRR .

Part A

What is the magnetic field in the middle of the capacitor plates at a distance \texttipdd from the center, as a function of time \texttiptt?

Give your answer in terms of \texttipdd, \texttipLL, \texttipI_\rm 0I_0, \texttiptt, \texttipRR, \texttipCC, and any necessaryconstants.

Hint 1. How to approach the problem

We are given a point on a circle in the middle of two capacitor plates, and we need to find the magnetic field at that point. Because we areworking with a highly symmetric shape, we can use Ampère's law to solve this problem. Draw a circle in the center of the capacitor passingthrough our point. This circle has radius \texttipdd. Apply Ampère's law to this circle.

Hint 2. What is the current in Ampère's law?

The current through the loop is not equal to the displacement current! The displacement current runs through the entire area of the capacitorplates, and the loop encloses only a fraction of that. You can find the current by taking the displacement current times the area of the loop overthe area of the plates (the fraction of the current that runs through the loop). That is, \largeB=\frac\mu_0 I_\rm encl2 \pi d. What is theequation for \texttipI_\rm enclI_encl?

Express the equation in terms of the radius of the circle \texttipdd, current \texttipII, and the radius of the capacitor plates \texttipLL.

ANSWER:

Hint 3. The displacement current of a charging capacitor

Recall how the current through a charging capacitor dies down with time. The current through a capacitor is equal to the initial current I_0 e^(-t/RC), where \texttiptt is the time, \texttipRR is the known resistance, and \texttipCC is the known capacitance. All you have todo is substitute this equation for the current, and you will have the answer.

ANSWER:

\cal E = = 1.32 \rm V

\texttipI_\rm enclI_encl =

Part B

Assume that the capacitor has been charging for a long, long time (t\to\infty). What is the approximate value of the magnetic field between the platesnow?

Hint 1. Current as t \rightarrow \infty

If the value for \texttiptt is very large, the value of \large\frac 1e^t must be very small. Therefore, you can approximate \large\frac 1e^t\approx 0.

ANSWER:

Induced Current in a Pair of Solenoids Conceptual QuestionDescription: Determine the current in a secondary coil induced by changes in current through a primary coil (or the equivalent).

For each of the actions depicted, determine the direction (right, left, or zero) of the current induced to flow through the resistor in the circuit containing thesecondary coil. The coils are wrapped around a plastic core.

Part A

Immediately after the switch is closed, as shown in the figure, in which direction does thecurrent flow through the resistor?

Hint 1. Magnetic flux

Magnetic flux is the product of the component of magnetic field perpendicular to a given area and the area itself. Conceptually, it is proportionalto the number of magnetic field lines passing through a given area.

Hint 2. Electromagnetic induction

Whenever magnetic flux through an area changes, an electromotive force (emf) is created around the area. This induced emf has a directionsuch that if a conductor is present, current will flow to create a secondary magnetic field that opposes the change in the original magnetic flux.Basically, the induced current will "try" to maintain the initial value of the magnetic flux.

Hint 3. Magnetic field produced by a coil

The magnetic field produced by a coil has the same directional properties as the field produced by a single loop. To determine the direction ofthe magnetic field produced by a current in a coil, use the right-hand rule for magnetic fields: Curl the finger of your right hand in the direction ofthe current flowing through the loop; your thumb now points in the direction of the magnetic field produced by the coil.

Hint 4. Find the direction of the magnetic field through the secondary coil

Initially, there is no magnetic field through the secondary coil (the coil connect to the resistor). After the switch is closed, current flows through

\texttipB\left(d,t\right)B(d,t) =

\texttipB\left(r,\infty\right)B(r,\infty) = 0 T

the primary coil and a magnetic field is produced through both coils. Does the magnetic field that passes through the secondary coil point to theright or to the left?

ANSWER:

Hint 5. Induced magnetic field

Since the magnetic field in the secondary coil changed from zero to a leftward-directed field when the switch was closed, the magnetic fluxthrough the secondary coil has increased. Thus, current will flow in the secondary coil to create a magnetic field oriented to the right to opposethis increase in flux. Therefore, you need to determine the direction in which current must flow through the secondary coil, and thus the resistor,to create a field through the secondary coil directed to the right. Use the right-hand rule for magnetic fields to find the direction of this current.

ANSWER:

Part B

If the switch is then opened, as shown in the figure, in which direction does the current flowthrough the resistor?

Hint 1. Find the direction of the magnetic field through the secondary coil before the switch is opened

Before the switch is opened, does the magnetic field that passes through the secondary coil point to the right or to the left?

Hint 1. Right-hand rule for magnetic fields

To find the direction of the magnetic field produced by a current flowing through a loop, curl the finger of your right hand in the direction ofthe current in the loop. Your thumb will point in the direction of the magnetic field produced by the coil.

ANSWER:

Hint 2. Induced magnetic field

Since the magnetic field in the secondary coil changed from a leftward-directed field to zero when the switch was opened, the magnetic fluxthrough the secondary coil had decreased. Thus, current will flow in the secondary loop to create a magnetic field oriented to the left to replacethis decrease in flux. Therefore, you need to determine the direction in which current must flow through the secondary coil, and thus the resistor,to create a field directed to the left. Use the right-hand rule for magnetic fields to find the direction of this current.

right

left

right

left

zero

right

left

ANSWER:

Part C

Immediately after the switch is closed, as shown in the figure, in which direction does thecurrent flow through the resistor?

ANSWER:

In Part D, the left coil (only) is moving to the left with velocity \texttip\vecvv_vec.

Part D

If the circuit containing the battery moves to the left, as shown in the figure, in which directiondoes the current flow through the resistor?

Hint 1. Find how the magnetic flux through the secondary changes

If the primary coil (the coil connected to the battery) moves to the left, how does the magnetic flux through the secondary coil change?

ANSWER:

right

left

zero

right

left

zero

It decreases.

It increases.

It remains constant.

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Hint 2. Induced magnetic field

Since the magnetic field in the secondary coil became weaker when the source of the field (the primary coil) moved away, the magnetic fluxthrough the secondary coil has decreased. Thus, current will flow in the secondary loop to create a magnetic field oriented in the same direction(to the left) as the original field to replace this decrease in flux. Therefore, you need to determine the direction in which current must flow throughthe secondary coil, and thus the resistor, to create a field to the left. Use the right-hand rule for magnetic fields to find the direction of thiscurrent.

ANSWER:

right

left

zero