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# Lecture 18 - Electromagnetic Induction - MAGNETIC FIELD Notes | EduRev

## : Lecture 18 - Electromagnetic Induction - MAGNETIC FIELD Notes | EduRev

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Module 3 : MAGNETIC FIELD
Lecture 18 : Electromagnetic Induction

Objectives

In this lecture you will learn the following
Study in detail the principle of electromagnetic induction.
Understand the significance of Lenz's law.
Solve problems involving motional emf.

Motional Emf :

Consider a straight conductor AB moving along the positive x-direction with a uniform speed . The region is in a
uniform magnetic field pointing into the plane of the page, i.e. in direction.
The fixed positive ions in the conductor are immobile. However, the negatively charged electrons experience a Lorentz
force , i.e. a force along the direction. This pushes the electrons from the end
A to the end B, making the former positive with respect to the latter. Thus an induced electric field is established in the
Page 2

Module 3 : MAGNETIC FIELD
Lecture 18 : Electromagnetic Induction

Objectives

In this lecture you will learn the following
Study in detail the principle of electromagnetic induction.
Understand the significance of Lenz's law.
Solve problems involving motional emf.

Motional Emf :

Consider a straight conductor AB moving along the positive x-direction with a uniform speed . The region is in a
uniform magnetic field pointing into the plane of the page, i.e. in direction.
The fixed positive ions in the conductor are immobile. However, the negatively charged electrons experience a Lorentz
force , i.e. a force along the direction. This pushes the electrons from the end
A to the end B, making the former positive with respect to the latter. Thus an induced electric field is established in the

conductor along the positive direction. The acceleration of electrons would stop when the electric field is built to a
strength which is strong enough to annul the magnetic force. This electric field is the origin of what is
known as motional emf . The motion of charges finally stops due to the resistance of the conductor
If the conductor slides along a stationary U- shaped conductor, the electrons find a path and a current is established in
the circuit. The moving conductor thereby becomes a seat of the motional emf. We may calculate the emf either by
considering the work that an external agency has to do to keep the sliding conductor move with a uniform velocity or
by direct application of Faraday's law.
If the induced current is , a force acts on the wire in the negative x direction. In order to maintain the uniform
velocity, an external agent has to exert an equal and opposite force on the sliding conductor. Since the distance moved
in time is , the work done by the external agency is
where is the amount of charge moved by the seat of emf along the direction of the current. The emf is an
electric potential difference.

Thus, the emf is equivalently the work done in moving a unit charge. Thus,
This emf corresponds to the potential difference beteen the ends A and B.
An alternate derivation of the above is to consier the flux linked with closed circuit. Taking the origin at the extreme
left end of the circuit, the area of the circuit in the magnetic field is where is the distance of the sliding rod from
the fixed end. The flux linked with the circuit is, therefore, . The rate at which flux changes is therefore
given by
If is not perpendicular to the plane of the circuit, we will need to take the perpendicular component of in the
above formula.
Note that, in the illustration above, the magnetic flux linked with the circuit is increasing with time in the negative z
direction. The direction of the induced current is, therefore, such that the magnetic field due to the current is along the
positve z-direction, which will oppose increase of flux. Hence the current, as seen from above, is in the anticlockwise
direction.

Example 18

In the above circuit if the part of the fixed rails parallel to the sliding conductor has a resistance and the rest of the
circuit may be considered resistanceless, obtain an expression for the velocity of the conductor at time , assuming
that it starts with an initial velocity at . Explain the change in the kinetic energy of the sliding rod.

Solution :

Since the emf is , the current flowing through the sliding conductor is .
Page 3

Module 3 : MAGNETIC FIELD
Lecture 18 : Electromagnetic Induction

Objectives

In this lecture you will learn the following
Study in detail the principle of electromagnetic induction.
Understand the significance of Lenz's law.
Solve problems involving motional emf.

Motional Emf :

Consider a straight conductor AB moving along the positive x-direction with a uniform speed . The region is in a
uniform magnetic field pointing into the plane of the page, i.e. in direction.
The fixed positive ions in the conductor are immobile. However, the negatively charged electrons experience a Lorentz
force , i.e. a force along the direction. This pushes the electrons from the end
A to the end B, making the former positive with respect to the latter. Thus an induced electric field is established in the

conductor along the positive direction. The acceleration of electrons would stop when the electric field is built to a
strength which is strong enough to annul the magnetic force. This electric field is the origin of what is
known as motional emf . The motion of charges finally stops due to the resistance of the conductor
If the conductor slides along a stationary U- shaped conductor, the electrons find a path and a current is established in
the circuit. The moving conductor thereby becomes a seat of the motional emf. We may calculate the emf either by
considering the work that an external agency has to do to keep the sliding conductor move with a uniform velocity or
by direct application of Faraday's law.
If the induced current is , a force acts on the wire in the negative x direction. In order to maintain the uniform
velocity, an external agent has to exert an equal and opposite force on the sliding conductor. Since the distance moved
in time is , the work done by the external agency is
where is the amount of charge moved by the seat of emf along the direction of the current. The emf is an
electric potential difference.

Thus, the emf is equivalently the work done in moving a unit charge. Thus,
This emf corresponds to the potential difference beteen the ends A and B.
An alternate derivation of the above is to consier the flux linked with closed circuit. Taking the origin at the extreme
left end of the circuit, the area of the circuit in the magnetic field is where is the distance of the sliding rod from
the fixed end. The flux linked with the circuit is, therefore, . The rate at which flux changes is therefore
given by
If is not perpendicular to the plane of the circuit, we will need to take the perpendicular component of in the
above formula.
Note that, in the illustration above, the magnetic flux linked with the circuit is increasing with time in the negative z
direction. The direction of the induced current is, therefore, such that the magnetic field due to the current is along the
positve z-direction, which will oppose increase of flux. Hence the current, as seen from above, is in the anticlockwise
direction.

Example 18

In the above circuit if the part of the fixed rails parallel to the sliding conductor has a resistance and the rest of the
circuit may be considered resistanceless, obtain an expression for the velocity of the conductor at time , assuming
that it starts with an initial velocity at . Explain the change in the kinetic energy of the sliding rod.

Solution :

Since the emf is , the current flowing through the sliding conductor is .

The force acting on the sliding conductor is
The force is a retarding one, slowing down the conductor in accordance with Lenz's law.
(This is the amount of force an external agency must apply on the sliding conductor in the positive x-direction to keep
the rod moving with a constant speed.) Thus the equation of motion of the conductor is
The equation may be solved by separating the variables and integrating from time to . We get
which gives

i.e.
We can calculate the power dissipated in the circuit in this time by using , so that
which is precisely the change in the kinetic energy of the conductor .
Page 4

Module 3 : MAGNETIC FIELD
Lecture 18 : Electromagnetic Induction

Objectives

In this lecture you will learn the following
Study in detail the principle of electromagnetic induction.
Understand the significance of Lenz's law.
Solve problems involving motional emf.

Motional Emf :

Consider a straight conductor AB moving along the positive x-direction with a uniform speed . The region is in a
uniform magnetic field pointing into the plane of the page, i.e. in direction.
The fixed positive ions in the conductor are immobile. However, the negatively charged electrons experience a Lorentz
force , i.e. a force along the direction. This pushes the electrons from the end
A to the end B, making the former positive with respect to the latter. Thus an induced electric field is established in the

conductor along the positive direction. The acceleration of electrons would stop when the electric field is built to a
strength which is strong enough to annul the magnetic force. This electric field is the origin of what is
known as motional emf . The motion of charges finally stops due to the resistance of the conductor
If the conductor slides along a stationary U- shaped conductor, the electrons find a path and a current is established in
the circuit. The moving conductor thereby becomes a seat of the motional emf. We may calculate the emf either by
considering the work that an external agency has to do to keep the sliding conductor move with a uniform velocity or
by direct application of Faraday's law.
If the induced current is , a force acts on the wire in the negative x direction. In order to maintain the uniform
velocity, an external agent has to exert an equal and opposite force on the sliding conductor. Since the distance moved
in time is , the work done by the external agency is
where is the amount of charge moved by the seat of emf along the direction of the current. The emf is an
electric potential difference.

Thus, the emf is equivalently the work done in moving a unit charge. Thus,
This emf corresponds to the potential difference beteen the ends A and B.
An alternate derivation of the above is to consier the flux linked with closed circuit. Taking the origin at the extreme
left end of the circuit, the area of the circuit in the magnetic field is where is the distance of the sliding rod from
the fixed end. The flux linked with the circuit is, therefore, . The rate at which flux changes is therefore
given by
If is not perpendicular to the plane of the circuit, we will need to take the perpendicular component of in the
above formula.
Note that, in the illustration above, the magnetic flux linked with the circuit is increasing with time in the negative z
direction. The direction of the induced current is, therefore, such that the magnetic field due to the current is along the
positve z-direction, which will oppose increase of flux. Hence the current, as seen from above, is in the anticlockwise
direction.

Example 18

In the above circuit if the part of the fixed rails parallel to the sliding conductor has a resistance and the rest of the
circuit may be considered resistanceless, obtain an expression for the velocity of the conductor at time , assuming
that it starts with an initial velocity at . Explain the change in the kinetic energy of the sliding rod.

Solution :

Since the emf is , the current flowing through the sliding conductor is .

The force acting on the sliding conductor is
The force is a retarding one, slowing down the conductor in accordance with Lenz's law.
(This is the amount of force an external agency must apply on the sliding conductor in the positive x-direction to keep
the rod moving with a constant speed.) Thus the equation of motion of the conductor is
The equation may be solved by separating the variables and integrating from time to . We get
which gives

i.e.
We can calculate the power dissipated in the circuit in this time by using , so that
which is precisely the change in the kinetic energy of the conductor .

Exercise 1

A pair of parallel conducting rails are inclined at an angle to the horizontal. The rails are connected to each other at
the ground by a conducting strip. A conductor of resistance , oriented parallel to the strip can slide down the incline
along the rails. The resistances of the rails and the strip are negligible.
A uniform magnetic field exists in the vertical direction. The slider is released at some height. Show that the slider
attains a terminal speed given by
where is the distance between the rails.
Exercise 2

In the above problem, show that after attaining the terminal velocity, the change in the potential energy of the slider is
equal to the Joule heat produced in the slider.
Hint for Solution :
In the preceding exercise, show that the current is given by . The rate of Joule heat is . The
amount of heat produced in time is . In time , the slider moves through a distance . The change in the
potential energy is .

Example 19
A conductor AB is moving with a speed parallel to a long straight wire carrying a current . What

is the potential difference between the ends A and B ?

Solution :

From the discussion above, the electric field acting at any point on the moving conductor is Lorentz force per unit
charge, i.e., . Since the field is not constant along AB, the emf is obtained by integrating the electric field
along AB.

The field at an element of width at a distance from the wire is . Since and are perpendicular,
Page 5

Module 3 : MAGNETIC FIELD
Lecture 18 : Electromagnetic Induction

Objectives

In this lecture you will learn the following
Study in detail the principle of electromagnetic induction.
Understand the significance of Lenz's law.
Solve problems involving motional emf.

Motional Emf :

Consider a straight conductor AB moving along the positive x-direction with a uniform speed . The region is in a
uniform magnetic field pointing into the plane of the page, i.e. in direction.
The fixed positive ions in the conductor are immobile. However, the negatively charged electrons experience a Lorentz
force , i.e. a force along the direction. This pushes the electrons from the end
A to the end B, making the former positive with respect to the latter. Thus an induced electric field is established in the

conductor along the positive direction. The acceleration of electrons would stop when the electric field is built to a
strength which is strong enough to annul the magnetic force. This electric field is the origin of what is
known as motional emf . The motion of charges finally stops due to the resistance of the conductor
If the conductor slides along a stationary U- shaped conductor, the electrons find a path and a current is established in
the circuit. The moving conductor thereby becomes a seat of the motional emf. We may calculate the emf either by
considering the work that an external agency has to do to keep the sliding conductor move with a uniform velocity or
by direct application of Faraday's law.
If the induced current is , a force acts on the wire in the negative x direction. In order to maintain the uniform
velocity, an external agent has to exert an equal and opposite force on the sliding conductor. Since the distance moved
in time is , the work done by the external agency is
where is the amount of charge moved by the seat of emf along the direction of the current. The emf is an
electric potential difference.

Thus, the emf is equivalently the work done in moving a unit charge. Thus,
This emf corresponds to the potential difference beteen the ends A and B.
An alternate derivation of the above is to consier the flux linked with closed circuit. Taking the origin at the extreme
left end of the circuit, the area of the circuit in the magnetic field is where is the distance of the sliding rod from
the fixed end. The flux linked with the circuit is, therefore, . The rate at which flux changes is therefore
given by
If is not perpendicular to the plane of the circuit, we will need to take the perpendicular component of in the
above formula.
Note that, in the illustration above, the magnetic flux linked with the circuit is increasing with time in the negative z
direction. The direction of the induced current is, therefore, such that the magnetic field due to the current is along the
positve z-direction, which will oppose increase of flux. Hence the current, as seen from above, is in the anticlockwise
direction.

Example 18

In the above circuit if the part of the fixed rails parallel to the sliding conductor has a resistance and the rest of the
circuit may be considered resistanceless, obtain an expression for the velocity of the conductor at time , assuming
that it starts with an initial velocity at . Explain the change in the kinetic energy of the sliding rod.

Solution :

Since the emf is , the current flowing through the sliding conductor is .

The force acting on the sliding conductor is
The force is a retarding one, slowing down the conductor in accordance with Lenz's law.
(This is the amount of force an external agency must apply on the sliding conductor in the positive x-direction to keep
the rod moving with a constant speed.) Thus the equation of motion of the conductor is
The equation may be solved by separating the variables and integrating from time to . We get
which gives

i.e.
We can calculate the power dissipated in the circuit in this time by using , so that
which is precisely the change in the kinetic energy of the conductor .

Exercise 1

A pair of parallel conducting rails are inclined at an angle to the horizontal. The rails are connected to each other at
the ground by a conducting strip. A conductor of resistance , oriented parallel to the strip can slide down the incline
along the rails. The resistances of the rails and the strip are negligible.
A uniform magnetic field exists in the vertical direction. The slider is released at some height. Show that the slider
attains a terminal speed given by
where is the distance between the rails.
Exercise 2

In the above problem, show that after attaining the terminal velocity, the change in the potential energy of the slider is
equal to the Joule heat produced in the slider.
Hint for Solution :
In the preceding exercise, show that the current is given by . The rate of Joule heat is . The
amount of heat produced in time is . In time , the slider moves through a distance . The change in the
potential energy is .

Example 19
A conductor AB is moving with a speed parallel to a long straight wire carrying a current . What

is the potential difference between the ends A and B ?

Solution :

From the discussion above, the electric field acting at any point on the moving conductor is Lorentz force per unit
charge, i.e., . Since the field is not constant along AB, the emf is obtained by integrating the electric field
along AB.

The field at an element of width at a distance from the wire is . Since and are perpendicular,

One can obtain the same expression by a direct application of Faraday's law. It is not necessary to have a physical
circuit to calculate the potential difference. One can imagine the conductor AB to be a rod sliding along a rail. As the
rod moves, the area of the closed region ACDB increases. We cansider an area element of width at a distance
from the wire. Since and the area vector are parallel,

Integrating,
Thus
since the rate of increase of the length of the rectangle is .
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