ROUTERA


Electromagnetic Waves

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
Question 1.

In a microwave oven, the food is kept in a plastic container and the microwave is directed towards the food. The food is cooked without melting or igniting the plastic container. Explain.


Answer:

The food which is being cooked contains water. The microwaves which are directed towards the food has a natural frequency that match with the frequency of water. The water molecules inside the food absorb energy from the microwaves and get heated up causing the food around it to heat up. Due to absorption of energy of the microwaves, the food gets cooked. Whereas the plastic does not have any water molecules which would absorb energy and the frequency also doesn’t match with the natural frequency of the microwaves so the plastic container remains unaffected.



Question 2.

A metal rod is placed along the axis of a solenoid carrying a high-frequency alternating current. It is found that the rod gets heated. Explain why the rod gets heated?


Answer:

The solenoid carrying a high frequency alternating current generates a continuously changing magnetic field across the rod placed along the axis. Due to this varying magnetic field across the rod, eddy currents are generated on the surface of the rod. The resistance of the rod and the eddy currents flowing on the rod are responsible for the generation of heat.



Question 3.

Can an electromagnetic wave be deflected by an electric field? By a magnetic field?


Answer:

Electromagnetic waves are made up of electric and magnetic fields but not charges. When an electromagnetic wave passes from a region of electric or magnetic field, it will show no deflection as no charge is present which would experience a force due to these fields and be deflected.



Question 4.

A wire carries an alternating current i = i0sinωt. Is there an

electric field in the vicinity of the wire?


Answer:

Given: Alternating current


The current is the rate of flow of charge per unit time. Mathematically


For a time, t, we can assume the charges in the wire are at rest,


then the charge will be given as



Similarly, a no. of charges can be assumed to be at rest. These


charges produce a time varying electric field.



Question 5.

A capacitor is connected to an alternating-currents source. Is there a magnetic field between the plates?


Answer:

Yes, there is a magnetic field between the capacitor plates.

According to Ampere Maxwell’s law, the magnetic field in a region is due to the current due to charge carriers and the changing electric field in a region. Mathematical form of the law is



where B is the magnetic field, dl is small element of length of an


Amperian loop, μ0 is the magnetic permeability of free space and its value is 4π × 10-7 T m A-1, Ienclosed is the current due to charge carriers (conduction current) and Id is the displacement current which


is related to changing electric field.


The displacement current is related to electric field as



where ϕE is the time varying electric flux through the surface


and ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2.


As the capacitor is charging, the charges at different time create a time varying electric field in the space between the capacitor plates.


This time varying electric field further creates an alternating electric flux though a surface present in between the plates.


The time varying electric flux is



According to Gauss’s law, the flux at time t is equal to 1/ϵ0 time the


charge enclosed at that time



The charge enclosed at any time t will be


where C is the capacitance of the capacitor and V0 is the amplitude of alternating voltage source and ω is the angular frequency of the source.


The magnetic field in between the capacitor plates is given by




Question 6.

Can an electromagnetic wave be polarized?


Answer:

Electromagnetic waves consist of oscillating electric and magnetic field which are perpendicular to each other and the direction of propagation as well. Thus electromagnetic waves are transverse in nature and can be polarized. Polarization of electromagnetic waves will restrict the vibrations of electric and magnetic field vectors in one direction only.



Question 7.

A plane electromagnetic wave is passing through a region. Consider the quantities

(a) electric field,

(b) magnetic field,

(c) electrical energy in a small volume and

(d) magnetic energy in a small volume.

Construct pairs of the quantities that oscillate with equal

frequencies.


Answer:

Explanation: Let us consider an electromagnetic wave traveling


in the z direction. The electric field E of the wave is assumed to be



where E0 is the amplitude of the electric field, k is the wave number


of the wave , λ is the wavelength, ω is the angular frequency


of the wave and t is the time. The frequency of the wave


is



Now the magnetic field can be considered as



where B0 is the amplitude of the magnetic field, k is the wave


number of the wave , λ is the wavelength, ω is the angular


frequency of the wave and t is the time. The frequency of


the wave is



The frequencies of electric and magnetic waves are same.


Hence (a) and (b) oscillate with the same frequency.


Now the electric field energy density of the electromagnetic wave is


given by the relation



where ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2 and Ude is the energy density of electric field. Now the energy of the wave over small volume V of the region is the product of energy density and the small volume V. The energy of the electric field is






The energy of the electric field is .


The angular frequency of the energy of the wave is 2ω, the


Corresponding frequency will be



Now the magnetic field energy density of the electromagnetic wave


is given by the relation



where μ0 is the magnetic permeability of free space and its value is


4π × 10-7 T m A-1, Udb is the magnetic field energy density. Now the energy of the wave over small volume V of the region is the product of energy density and the small volume V. The energy of the magnetic field is






The energy of the magnetic field is


The angular frequency of the energy of the wave is 2ω, the


Corresponding frequency will be



The electric field energy and magnetic field energy have the same


frequencies of oscillation.


So (c) and (d) form a pair.



Objective I
Question 1.

A magnetic field can be produced by
A. a moving charge

B. a changing electric field

C. none of them

D. both of them


Answer:

Explanation: A magnetic current can be produced by a moving


charge. If the charge is moving uniformly, it will generate a constant


current. If we assume a constant current is flowing through a


conductor of length L, the magnetic field due to it at a distance R


from it is given by Biot-Savart law. The field would be



where i is the constant current and μ0 is the magnetic permeability


of free space and its value is 4π × 10-7 T m A-1. So A. is correct.


A changing electric field can also generate a magnetic field around it. According to Ampere-Maxwell law, the total magnetic field around a region is due to flow of charge carriers (called as conduction current)


and the time varying electric flux. Mathematically


…(i)


where B is the magnetic field, dl is small element of length of an


Amperian loop, μ0 is the magnetic permeability of free space and its value is 4π × 10-7 T m A-1, Ienclosed is the current due to charge carriers (conduction current) and Id is the displacement current which


is related to changing electric field.


The displacement current is related to electric field as


…(ii)


where ϕE is the time varying electric flux through the surface


and ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2.


The electric flux is the amount of electric field lines passing


through a surface normally. Mathematically


…. (iii)


where E(t) is the time varying electric field, dS are a small area


element on the surface and θ is the angle between the electric field


vector and area vector (if the field lines are not falling normally).


From (ii) and (iii)



If the conduction current is zero, so the magnetic field will entirely be


produced by the changing electric field and would be produced by it.


Hence B. is correct


Hence D. is correct


Question 2.

A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle
A. does not deflect

B. deflects for a very short time and then comes back to the original position

C. deflects and remains deflected as long as the battery is connected.

D. deflects and gradually comes to the original position in a time which is large compared to the time constant.


Answer:

deflects and gradually comes to the original position


in a time, which is large compared to the time constant.


Explanation: The deflection in a compass needle are because of


magnetic field around the region of the compass. The magnetic field


arises due to charging of the capacitor which creates a varying


electric field. This varying electric field produces a changing electric


field through the capacitor plates. Due to this changing electric flux, a current called the displacement current arises in the capacitor. The


current Id is related to the electric flux as


…. (i)


where ϕE is the time varying electric flux through the plane surface


and ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2.


According to Ampere-Maxwell’s law of electromagnetism, the total


magnetic field is due to current due to charge carriers and the


current due to time varying electric flux. Ampere-Maxwell law is the


extension of Ampere Circuital Law. The mathematical relation


is



…(ii)


where B is the magnetic field, dl is small element of length of an


Amperian loop, μ0 is the magnetic permeability of free space and its value is 4π × 10-7 T m A-1, and Ienclosed is the current due to charge carriers.


From (i) and (ii)


…(iii)


According to Gauss’s Law of electrostatics, the electric flux through


a surface is equal to 1/ϵ0 times the charge enclosed by it.



The charge enclosed by the capacitor plates is changing with time


as the capacitor is charging. The charge is given as



where


C=capacitance of the capacitor


V=potential of the battery


R=resistance of the resistor connected in series


Using the above two relations in (iii)



So the magnetic field depends on the charge of the capacitor. The


needle will show deflection till the capacitor gets charged to a


specific value at a time τ called the time constant. is the time


constant of the capacitor. Charging after this time will be gradual so


the needle will come to its original position.


Question 3.

Dimensions of 1/(μ0ϵ0) is
A. L/T

B. T/L

C. L2/T2

D. T2/L2.


Answer:

Explanation: The electrical permittivity of free space and magnetic


permeability of free space is related with the speed of light in


vacuum as



squaring the above relation



The dimension of will be the dimension of C2.


The dimension of C is , so the dimension of C2 will be



Therefore C. is the only correct option.


Question 4.

Electromagnetic waves are produced by
A. a static charge

B. a moving charge

C. an accelerating charge

D. charge less particles


Answer:

Explanation: An accelerating charge produces a changing electric


field which in turn produces a magnetic field. These alternatively


changing magnetic and electric fields give rise to electromagnetic


waves. A static charge gives rise to only an electric field and a


moving charge creates a magnetic field so A. and B. are incorrect.


Question 5.

An electromagnetic wave going through vacuum is described by

E = E0 sin(κx – ωt); B = B0 sin(κx – ωt)

Then

A. E0 κ = B0 ω

B. E0 B0 = ωκ

C. E0 ω = B0 κ

D. None of these


Answer:

Explanation: The amplitudes of the electric and magnetic fields are


related as


…(i)


where c is the speed of wave and E0 and B0 are the amplitudes of the fields.


The speed of wave is related with its frequency and wavenumber as


…(ii)


where ω is the angular frequency of the wave ( and k is the wavenumber () and ν is the frequency and λ is the wavelength of the wave.


From (i) and (ii)



so A. is correct.


B. and C. are invalid relations. Only A. is the correct option.


Question 6.

An electric field and a magnetic field exist in a region. The fields are not perpendicular to each other.
A. This is not possible

B. No electromagnetic wave is passing through the region.

C. An electromagnetic wave may be passing through the region.

D. An electromagnetic wave is certainly passing through the region.


Answer:

Explanation: There exists a mode of propagation of electromagnetic wave called the Transverse Electric and Magnetic (TEM) mode where both electric and magnetic field are moving tranverse to the direction of propagation of wave. The fields are not in the direction of propagation.


It is possible that an electromagnetic wave may exist when electric and magnetic fields are not perpendicular to each other.



Question 7.

Consider the following two statements regarding a linearly polarized, plane electromagnetic wave:

(A) The electric field and the magnetic field have equal average values.

(B) The electric energy and the magnetic energy have equal average values.

A. Both A and B are true

B. A is false but B is true

C. B is false but A is true

D. Both A and B are false


Answer:

Explanation: Let us assume the linearly polarized plan wave travels


in z direction so the electric and magnetic fields of the wave are


given by




Their average values over one cycle would be the average from time


t=0 to t=T.


Average of electric field






(because and )


Therefore, the average of electric field is zero.


Similarly average of magnetic field is also zero as






So electric and magnetic field have their average value zero.


Therefore (A) is correct.


Now the electric energy associated with the electric field is given by


…(i)


where ϵ0 is the electric permittivity of free space(vacuum) and is


equal to 8.85 × 10-12 C2 N-1 m-2 and Ude is the energy density of


electric field.


Similarly, the magnetic energy associated with the magnetic field is


given by


…(ii)


where μ0 is the magnetic permeability of free space and its value is


4π × 10-7 T m A-1, Udb is the magnetic field energy density.


Also the amplitudes of electric and magnetic fields are related as


…(iii)


where C is the speed of light in vacuum.


and


Using (iii) in (i)


…. (iv)


The electrical permittivity and magnetic permeability of free space


related as


…. (v)


Using (v) in (iv)




Therefore, the electric and magnetic energy densities are same in


magnitude and so their average values will be same.


Average of electric energy over one cycle will be







So electric and magnetic energy and have non zero average values.


Therefore (B) is also correct.


Question 8.

A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving
A. along the electric field

B. along the magnetic field

C. along the direction of propagation of the wave

D. in a plane containing the magnetic field and the direction of propagation.


Answer:

Explanation: A particle placed in and electromagnetic wave moving


with velocity v experiences a force called Lorentz’s force due to the


electric and magnetic fields. The force is given as



where q is the charge of the particle, E is the electric field, v is the


velocity of the particle and B is the magnetic field.


An electron at rest is placed in the path of a plane electromagnetic wave. The Lorentz force acting on the electron will be



where e is the charge of the electron. As the electron is at rest so the velocity will be 0. The net Lorentz’s force will be



The electron will move in the direction of the field.


Question 9.

A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.
A. p = 0, E ≠ 0

B. p ≠ 0, E = 0

C. p ≠ 0, E ≠ 0

D. p = 0, E = 0


Answer:

Explanation: An electromagnetic wave striking a material surface


delivers both energy and momentum to the surface.


If the frequency of the wave is ν then the energy imparted will be


…. (i)


where h is the Planck’s constant.


The energy is also related to the speed of the wave as


…. (ii)


where p is the momentum and c is the speed of the wave.


From (i) and (ii)




The energy and momentum both are non zero.


We can also define a quantity known as radiation pressure which is the pressure the radiation exerts on the material surface is strikes.


The radiation pressure is given as


…. (iii)


Also the electromagnetic wave has energy so there is an intensity related to it given by


…. (iv)


From (iii) and (iv)



(because energy dissipated per unit time is the power and power = force × velocity)


So is the radiation pressure.



Objective Ii
Question 1.

An electromagnetic wave going through vacuum is described by

E = E0sin (κx – ωt).

Which of the following is/are independent of the wavelength?

A. κ

B. ω

C. κ/ω

D. κω


Answer:

Explanation:


Given: The equation of an electromagnetic wave is given as



where E0 is the amplitude of the electric field, k is the wave number


of the wave , λ is the wavelength, ω is the angular frequency


of the wave and t is the time. The frequency of the wave


is .


A. is k which is equal to and is dependent on λ so A. is


incorrect.


B is ω which is . If the speed of the wave is c, then


, so ω becomes , hence B. in incorrect.


C. is , substituting the values of ω and k we get =which is


independent of wavelength so C. in correct.


D. is kω, using the values of k and ω, which is


dependent on wavelength so D. is in incorrect.


Question 2.

Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor
A. increases

B. decreases

C. does not change

D. is zero


Answer:

Explanation: The displacement current depends on the changing


electric flux across the capacitor plates. Due to the charging of the


capacitor, the charge varies on the capacitor plates with time. Due


to this, the electric flux also varies with time. The relation between


the displacement current and the electric flux is given as



where ϕE is the time varying electric flux through the plane surface,


ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2 and Id is the displacement current.


The electric flux is given by Gauss’s law as



The displacement current then becomes




which is dependent on charge. If the charge is zero, then the R.H.S


would be zero so D. is incorrect. If the charge doesn’t change with


time so its derivative w.r.t time will be zero and so will be the


current, so C. is zero. For R.H.S to be non-zero, the charge must vary with time i.e. increase or decrease, hence A. and B. are correct.


Question 3.

Speed of electromagnetic waves is the same
A. for all wavelengths

B. in all media

C. for all intensities

D. for all frequencies


Answer:

Explanation: The speed of electromagnetic waves is given by



where ν is the frequency and λ is the wavelength. Thus the speed


will be not same for all frequencies and wavelengths so A. and D.


are incorrect. Also as wave travels from one medium to another, the


speed changes so B. is incorrect. The relation between speed and


intensity of a wave is given by



where ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2, c is the speed of wave and E0 is the amplitude of electric field. The speed will be same for all intensities in a given medium will be same so C. is correct.


Question 4.

Which of the following have zero average value in a plane electromagnetic wave?
A. electric field

B. magnetic field

C. electric energy

D. magnetic energy


Answer:

Explanation: Let us assume the wave travels in z direction so the


electric and magnetic fields of the wave are given by




Their average values over one cycle would be the average from time


t=0 to t=T.


Average of electric field






(because and )


Therefore, the average of electric field is zero.


Similarly average of magnetic field is also zero as






So electric and magnetic field have their average value zero.


Now the electric energy associated with the electric field is given by


…(i)


where ϵ0 is the electric permittivity of free space(vacuum) and is


equal to 8.85 × 10-12 C2 N-1 m-2 and Ude is the energy density of


electric field.


Similarly, the magnetic energy associated with the magnetic field is


given by


…(ii)


where μ0 is the magnetic permeability of free space and its value is


4π × 10-7 T m A-1, Udb is the magnetic field energy density.


Also the amplitudes of electric and magnetic fields are related as


…(iii)


where C is the speed of light in vacuum.


and


Using (iii) in (i)


…. (iv)


The electrical permittivity and magnetic permeability of free space


related as


…. (v)


Using (v) in (iv)




Therefore, the electric and magnetic energy densities are same in


magnitude and so their average values will be same.


Average of electric energy over one cycle will be







So electric and magnetic energy and have non zero average values.


Question 5.

The energy contained in a small volume through which an electromagnetic wave is passing oscillates with
A. zero frequency

B. the frequency of the wave

C. half the frequency of the wave

D. double the frequency of the wave


Answer:

Explanation: Let us consider an electromagnetic wave traveling


in the z direction. The electric field E of the wave is assumed to be



where E0 is the amplitude of the electric field, k is the wave number


of the wave , λ is the wavelength, ω is the angular frequency


of the wave and t is the time. The frequency of the wave


is



Now the electric field energy density of the electromagnetic wave is


given by the relation



where ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2 and Ude is the energy density of electric field. Now the energy of the wave over volume V of the region is the product of energy density and the volume V. The energy of the electric field is






The energy of the electric field is .


The angular frequency of the energy of the wave is 2ω, the


Corresponding frequency will be



This frequency f’ is twice of f, .


Now the magnetic field can be considered as



where B0 is the amplitude of the magnetic field, k is the wave


number of the wave , λ is the wavelength, ω is the angular


frequency of the wave and t is the time. The frequency of


the wave is



Now the magnetic field energy density of the electromagnetic wave


is given by the relation



where μ0 is the magnetic permeability of free space and its value is


4π × 10-7 T m A-1, Udb is the magnetic field energy density. Now the energy of the wave over volume V of the region is the product of energy density and the volume V. The energy of the magnetic field is






The energy of the magnetic field is


The angular frequency of the energy of the wave is 2ω, the


Corresponding frequency will be



This frequency f’ is twice of f, .


As for both, electric and magnetic, the energy of wave is double the


frequency of their respective waves, so the overall energy of the


wave has twice the frequency of oscillation of the wave itself.



Exercises
Question 1.

Show that the dimensions of the displacement current are that of an electric current.


Answer:

The displacement current Id is produced by a varying electric


field. It is given by the relation



where ϵ0 is the electric permittivity of free space(vacuum) and is


equal to 8.85 × 10-12 C2 N-1 m-2 and ϕE is the electric flux produced


by the time varying electric field.


To find the dimension of Id and check whether it is the same as that


of electric current, we need to find the dimension of .


We can simplify using Gauss’s Law. According to Gauss’s law


the electric flux ϕE through a surface is given as



Using the above relation, the displacement current becomes




The dimension of displacement current is the dimension of the


quantity in the R.H.S of the above equation



As [A] is the dimension of electric current so displacement current


has same dimension as that of electric current.



Question 2.

A point charge is moving along a straight line with a constant

velocity v. Consider a small area A perpendicular to the direction of motion of the charge figure. Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large so that the electric field at any instant is essentially given by Coulomb’s law.




Answer:

Given: Velocity of charge = v


Area of patch = A


Distance of charge from the patch=x


We have to find the displacement current through the area when it is at a distance x from the charge.


The displacement current arises due to changing electric field which in turn produces a varying electric flux through an area. The displacement current depends on the electric flux as



where Id is the displacement current,ϕE is the varying electric flux


through the area and ϵ0 is the electric permittivity of free


space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2.


The electric field produced by the charge when it is at a distance x


from the surface is given by Coulumb’s law and is equal to



This electric field produces an electric flux through the area whose


magnitude is given by Gauss’s law



. This is because the electric field lines are directed along the normal to the area vector of the surface.


The angle θ between them is 0° so cos 0° =1.



The displacement current will be





As x is the distance of the charge from the area at different intervals of time so the rate at which the particle is changing its position is its velocity v



Thus the displacement current through the area is



Question 3.

A parallel-plate capacitor having plate-area A and plate separation

d is joined to a battery of emf ϵ and internal resistance R at t = 0.

Consider a plane surface of area A/2, parallel to the plates and

situated symmetrically between them. Find the displacement current

through this surface as a function of time.


Answer:

Given: Area of capacitor plates=A


Separation between the plates=d


Emf of the battery = ϵ


Internal resistance of the battery = R


Area of plane surface= A/2


Displacement current is the current which is generated by a time


varying electric field, not by the flow of charge carriers.


This current is also responsible for the generation of a time varying


magnetic field. The displacement current Id is generated due to the


fact that the charge on capacitor plates is changing with time.


The displacement current is given by



where ϕE is the time varying electric flux through the plane surface


and ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2.


The electric field in the space between the plates can be given by


Guass’s Law. If the charge on the capacitor plate is Q and the area


of the plate is A(given), then by Guass’s law,



where E is the electric field and ϵ0 is the electric permittivity of free


space and dS is a small area element on the plate.


Further (because the area vector


and electric field lines are both normal to the surface and in


same direction i.e. θ=0° so cos θ=1)


So , the electric field between the plates is .


This electric field produces and electric flux through the plane


surface given by



(because the area vector and electric field lines are both normal to


the surface and in same direction i.e. θ=0° so cos θ=1)



Now the charge on the capacitor is changing with time as it is


charging. If the capacitance of the capacitor is C, then the charge Q


at time t will be


where ϵ is the potential between plates which is equal to the emf of battery and R is the resistance attached in series.


The displacement current Id is given as






Thus the displacement current as a function of time is .



Question 4.

Consider the situation of the previous problem. Define displacement resistance Rd = V/id of the space between the plates where V is the potential difference between the plates and id is the displacement current. Show that Rd varies with time as

Rd = R (et/τ – 1)


Answer:

Given: The displacement resistance


We will first calculate the displacement current.


The displacement current Id is generated due to the fact that the


charge on capacitor plates is changing with time.


The displacement current is given by



where ϕE is the time varying electric flux through the plane surface


and ϵ0 is the electric permittivity of free space(vacuum) and is equal


to 8.85 × 10-12 C2 N-1 m-2.


We need to calculate the electric flux through the capacitor plate.


As the charge on the capacitor plate at time t can be taken as Q so


by using Gauss’s law, we will calculate the electric flux.


According to Gauss’s law



so the electric flux would be


.


As the capacitor is charging, the charge will be a function of time


given as



where C is the capacitance of the capacitor, V is the potential drop


at time t, R is the series resistance and V0 is the potential at time


t=0. Now the flux is .


Now by our definition, the displacement current is given by



which is




The displacement current is


We know that



therefore , where τ is the time constant.



Question 5.

Using B = μ0 H find the ratio E0/H0 for a plane electromagnetic wave propagating through vacuum. Show that is has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.


Answer:

Given: The magnetic field density and the magnetic field strength


as related by the relation



where B is the magnetic field density, μm is the relative magnetic


permeability of the material and depends on the nature of the


material or the medium and H is the magnetic field strength which


tells upto what extent a material can be magnetized.


For free space, the relation becomes


…. (i)


where B0 is the field intensity in free space(vacuum), H0 is the


strength in free space and μ0 is the magnetic permeability of free


space. μ0 is a universal constant and its value is 4π × 10-7 T m A-1.


We can define a relation between the electric field and magnetic field


of an electromagnetic wave travelling in vacuum. It is


…. (ii)


where E0 is the electric field density and C is the speed of light in


vacuum.


From (i) and (ii)




…. (iii)


To find the dimension of , we will find the dimension of the


quantity in the R.H.S.


The dimension of C is [L T-1] because it is the speed.


To find the dimension of , we need to consider the Biot Savart’s


law which gives the magnetic field due to a current carrying


conductor. According to Biot Savart law


…. (iv)


where i is the current in the conductor, L is the length of conductor and R is the distance between conductor and point where field is to be found.


The scalar form of the above equation will be simply


…. (v)


From (v)



the dimensions of μ0 will be the dimensions of R.H.S


dimension of B= [M T-2 A �-1]


dimension of R2=[L2]


dimension of L=[L]


dimension of i=[A]


dimension of R.H.S


Therefore dimension of μ0 is .


Now the dimension of R.H.S of eq(iii) will be dimension of C × dimension of μ0


dimension of


As the dimensions of electric resistance given by Ohm’s Law(V=iR) is



Therefore has the same dimensions as that of electrical resistance.


Also the value of is a constant because the R.H.S of eq(iii) is the


product of 2 universal constants.



Question 6.

The sunlight reaching the earth has maximum electric field of 810 V m–1. What is the maximum magnetic field in this light?


Answer:

Given: Maximum electric field E0 = 810 V m-1.


The sunlight travels through vacuum in outer space to reach the earth. The magnetic field and electric field are related to each other and their relation is



where E0 is the amplitude of the electric field, C is the speed of light in vacuum and B0 is the amplitude of maximum magnetic field.


Thus the maximum magnetic field is



The maximum magnetic field is 2.7 μT.



Question 7.

The magnetic field in a plane electromagnetic wave is given by

B = (200 μT) sin [(4.0 × 1015 s–1) (t –x/c)].

Find the maximum electric field and the average energy density corresponding to the electric field.


Answer:

Given: The equation of magnetic field of a plane electromagnetic


wave


B = (200 μT) sin [(4.0 × 1015 s–1) (t –x/c)]


where the amplitude of magnetic field is .


The amplitude of the electric field is related with amplitude of


magnetic field as



where C is the speed of light in free space and E0 is the amplitude


of electric field.


Thus the electric field intensity is given as





The energy density associated with an electric field is given as



where Ud is the energy density, ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2 and E0 is the amplitude of electric field.


Thus the electric field energy density is given by






The maximum electric field is and the corresponding


electric field energy density is .



Question 8.

A laser beam has intensity 2.5 × 1014 W m–2. Find the amplitudes of electric and magnetic fields in the beam.


Answer:

Given: The intensity of laser beam =2.5 × 1014W m-2.


We have to find the amplitudes of electric and magnetic field. The


amplitude of the electric field is related to the intensity of the wave


by the relation



where


ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2, C is the speed of light in vacuum and is equal to 3 × 108 m s-1 and E0 is the amplitude of electric field.





.


The amplitudes of electric and magnetic fields are related as



where B0 is the amplitude of magnetic field in free space.


Thus the value of magnetic field will be


.


The electric field amplitude is and the magnetic field amplitude is .



Question 9.

The intensity of the sunlight reaching Earth is 1380 W m−2. Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.


Answer:

Given: The intensity of sunlight reaching the earth = 1380 W m-2.


We have to find the amplitudes of electric and magnetic field if the


light is a plane electromagnetic wave. The amplitude of the electric


field is related to the intensity of the wave by the relation



where


ϵ0 is the electric permittivity of free space(vacuum) and is equal to 8.85 × 10-12 C2 N-1 m-2


C is the speed of light in vacuum and is equal to 3 × 108 m s-1


and E0 is the amplitude of electric field.






The amplitudes of electric and magnetic fields in free space are related as



where B0 is the amplitude of magnetic field in free space.


Thus the value of magnetic field will be



The electric field amplitude is and the magnetic field amplitude is.