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Q # 1. The potential is constant throughout a given region of space. Is the electric field zero or non zero in

this region? Explain.

Ans. The electric field intensity is described by the relation:

According to the relation, the electric field is negative gradient of electric potential. If the electric

potential is constant throughout given region of space, then change in electric potential , hence .

Q # 2. Suppose that you follow an electric field line due to a positive point charge. Do electric field and the

potential increases or decreases.

Ans. If we follow an electric field line due to a positive point charge, then it means that we are moving await

from point charge. Thus the distance from the charge increases. Due to increase of distance from positive

charge, both electric field intensity and electric potential decreases as:

and

Q # 3. How can you identify that which plate of capacitor is positively charged?

Ans. The presence of charge on a body is detected by a device called gold leaf electroscope. The leaves of gold

leaf electroscope are diverged by giving them negative charge.

Ø If the disc is touched with any plate of the charged capacitor and the divergence of the leaves increases,

the plate of capacitor is negatively charged

Ø If the divergence of leaves decreases, then that plate of capacitor is positively charged.

Q # 4. Describe the force or forces on a positive point charge when placed between parallel plates:

i. With similar and equal charges

ii. With opposite and equal charges

Ans. When a positive point charge is placed between parallel plates with similar and equal charges, then the

electric field intensity due to one plate is equal in magnitude but opposite in direction of electric intensity due to

other plate. So the value of resultant electric field intensity E is zero. Hence the net force on the positive point

charge is zero. Thus it will remain at rest.

When a positive point charge is placed between parallel plates with opposite but equal amount of

charge, then electric field intensity due to one plate is equal in magnitude but in same direction of the electric

field intensity due to other plate. So the value of resultant electric field intensity is non zero. Hence the point

charge will be accelerated towards negative plate.

Q # 5. Electric lines of force never cross. Why?

Electric lines of force never cross each other. This is because of the reason that electric field intensity has only

one direction at any given pint. If the lines cross, electric intensity could have more than one direction which is

physically not correct.

Q # 6. If a point charge of mass m is released in a non-uniform electric field with

field lines in the same direction pointing, will it make a rectilinear motion.

Ans. A non-uniform field of a positive point charge is shown in the figure:

If a point charge q of mass m is placed at any point in the field, it will follow

straight or rectilinear path along the field line due to repulsive force.

Q # 8. Is it true that Gauss’s law states that the total number of lines of force crossing any closed surface

in the outward direction is proportional to the net positive charge enclosed within surface?

Ans. Yes, the above statement is true.

Electric flux is defined as the measure of number of electric lines of force passing through a certain

area. According to Gauss‟s law, the flux through any close surface is

times the total charged enclosed in it.

Electric flux =

(Total Charge Enclosed)

Electric flux = constant (Total Charge Enclosed)

Electric flux (Total Charge Enclosed)

Q # 9. Do electrons tends to go to region of high potential or of low potential?

Ans. The electrons being negatively charge particle when released in electric field moves from a region of lower

potential (negative end) to a region of high potential (positive end).

Q # 1. A plane conducting loop is located in a uniform magnetic field that is directed along the x-axis. For

what orientations of the loop, is the flux maximum? For what orientation, is the flux minimum?

Ans. The magnetic flux through a conducting loop can be find out by the expression:

Here B is the magnetic field strength and A is vector area whose direction is perpendicular to the plane

of the loop.

Case 1. When vector area of the conducting loop is in the direction of magnetic field strength i.e., , then

the magnetic flux:

as

Thus the magnetic flux through the coil is maximum, when the vector area of the conducting loop is parallel to

magnetic field strength.

Case 2. When vector area of the conducting loop is perpendicular to magnetic field strength i.e., , then

the magnetic flux:

as

Thus the magnetic flux through the coil is minimum, when the vector area of the conducting loop is

perpendicular to magnetic field strength.

Q # 2. A current in a conductor produce a magnetic field, which can be calculated using Ampere’s Law.

Since current is defined as the rate of flow of charge. What can you conclude about the magnetic field due

to stationary charges? What about moving charges?

Ans. A stationary charges cannot produce any magnetic field. In case of stationary charges, the rate of flow of

charges is zero( i.e. current = 0), so there will be no magnetic field.

As the moving charges produce current, so the magnetic field produced around the path of its motion

similar to the magnetic field produced around a current carrying conductor.

Q # 3. Describe the charge in the magnetic field inside a solenoid carrying steady current I, if (a) the

length of the solenoid is doubled but the number of turns remains the same and (b) the number of turns

are doubled, but the length remains the same.

Ans. The magnetic field strength B inside a current carrying conductor can be find out by the expression:

———– (1)

Where I is the current flowing through conductor and n is the number of turns per unit length i.e.,

. Thus

(a) When Length of solenoid is doubled by keeping the number of turns constant, then magnetic field strength:

Thus on doubling the length of solenoid by keeping the turns constant, the magnetic field strength becomes one

half of its original value.

(b) When number of turns of solenoid is doubled by keeping the length of solenoid constant, then magnetic

field strength:

( )

Thus on doubling the number of turns of solenoid by keeping its length constant, the magnetic field strength

becomes doubled of its original value.

Q # 4. At a given instant, a proton moves in the positive x-direction in the region where there is magnetic

field in the negative z-direction. What is the direction of the magnetic force? Will the proton continue to

move in the positive x-direction? Explain.

Ans. As the proton is moving in the positive x-direction and magnetic field is directed into the plane of paper,

then the magnetic force on proton can be find out using expression:

( )

According to right hand rule, the magnetic force is directed along y-axis.

No, the proton will not continue to move in the positive x-direction. Since the magnetic force is acting

at the right angle to motion of conductor, therefore it will move along a circular path in xy-plane.

Q # 5. Two charged particles are projected into a region where there is a magnetic field perpendicular to

their velocities. If the charge are deflected in opposite directions, what can you say about them?

Ans. When a charge particle is projected in a magnetic field, it will experience the magnetic force given by:

( )

The magnetic force is a deflecting force. Thus if the charged particles are deflected in opposite direction, then

particles are oppositely charged. i.e., one particle is positively charged and the other is negatively charged.

Q # 6. Suppose that a charge q is moving in a uniform magnetic field with a velocity v. Why is there no

work done by the magnetic force that acts on the charge?

Ans. The magnetic force on a charge particle will act normal to the direction of motion of the particle, so the

work done by the force is given by:

Where is the angle between the magnetic force and displacement of charge particle. For present case:

. Therefore:

Thus we can say that magnetic force is a deflecting force and it cannot do any work.

Q # 7. If a charge particle moves in a straight line through some region of space, can you say that the

magnetic field in the region is zero.

Ans. The magnitude of magnetic force on a charge particle can be expressed as:

Where is the angle between B and v. So if the particle moves in a straight line through some region of space

then it means that the charge particle is not experiencing magnetic force which might be due to one of the

following reasons:

i. Magnetic field strength B in the region is zero

ii. Magnetic field is parallel or anti-parallel to the direction of motion.

Q # 8. Why does the picture on a TV screen become distorted when a magnet is brought near the screen?

Ans. The picture on a TV is formed when moving electrons strike the florescent screen. As magnet is brought

close to the TV screen, the path of electrons is distorted due to the magnetic force on them. So the picture on the

screen of TV is distorted.

Q # 9. Is it possible to orient a current loop in a uniform magnetic field such that the loop will not tend to

rotate? Explain.

Ans. A current carrying loop when placed in magnetic field will experience a torque given by:

Where B is the magnetic field strength, I is current flowing through coil, N is number of turns in a coil, A is the

area of the coil and is the angle between plane of the coil and magnetic field.

It is clear from expression that when plane of the coil makes and angle of with magnetic field, the

torque on the coil will be zero. In this condition, the coil will not tend to rotate.

Q # 10. How can a current loop be used to determine the presence of a magnetic field in a given region of

space?

Ans. When a current carrying loop is placed in a uniform magnetic field, a torque is produced in the loop is

given by:

If the loop is deflected in a given region, then it confirms the presence of magnetic field, otherwise not.

Q # 11. How can you use a magnetic field to separate isotopes of chemical element?

Ans. If the ions of isotopes of an element are projected in a magnetic field of known strength B, the ions move

in circular path of radius r. The e/m of the ion is given by the expression:

If v, B and e of the ions are constant, then

So the ions of different mass will have different radii of curvature and hence they can be separated in magnetic

field.

Q # 12. What should be the orientation of a current carrying coil in a magnetic field so that torque acting

upon the coil is (a) maximum (b) minimum?

Ans. A current carrying loop when placed in magnetic field will experience a torque given by:

Where B is the magnetic field strength, I is current flowing through coil, N is number of turns in a coil, A is the

area of the coil and is the angle between plane of the coil and magnetic field.

(a) When plane of the coil is parallel to magnetic field, and the torque acting on the coil will be

maximum given by:

(b) When plane of the coil is perpendicular to magnetic field, and the torque acting on the coil

will be minimum, given by: .

Q # 13. A loop of wire is suspended between the pools of a magnet with its plane parallel to the pole faces.

What happens if a direct current is put through the coil? What happens if an alternating current is used

Ans. When direct current is passed through he coil of wire, a torque acts on the coil which rotates the coil.

Chapter 14 (2nd Year Physics) Electromagnetism

Q # 14. Why the resistance of an ammeter should be very low?

Ans. An ammeter is connected in series with a circuit to measure the current. It is connected in series so that

total current passing through the circuit should pass through it. If the resistance of the ammeter will be large, it

will alter the current of the circuit to great extent and the measurement of current will not be accurate.

Q # 15. Why the voltmeter should have a very high resistance?

Ans. A voltmeter is connected in parallel to the resistor to measure potential difference across it. It should have

very high resistance so that practically, a very little current should pass through it and the current of the circuit

should almost remain constant, so that it might measure the potential difference across a resistor accurately.

Q # 1. Does the inducted emf in circuit depend on the resistance of the circuit? Does the induced

current depend on the resistance of the circuit?

Ans. The expression for induced emf is given by

The relation shows that the induced emf in a coil only depend upon the rate of change of

magnetic flux and number of turns but does not depend upon the resistance of the coil.

As the induced current flowing through a coil is given by:

this expression shows that the value of current depends upon the resistance of the coil. The

smaller the value of the resistance of the coil, greater will he the value of current.

Q # 2. A square loop of wire is moving through a uniform magnetic field. The normal to the loop

is oriented parallel to the magnetic field. Is a emf induced in the loop? Give a reason for your

Ans. The induce emf in a wire is given by:

Where the angle between “ ” and “ ”.

When normal to the loop is parallel to the field, the velocity vector “ ” of

side of loop is also parallel to field “ ”, so . Therefore,

( )

Thus, emf induced in the loop is zero.

Q # 3. A light metallic ring is released from above into a vertical bar

magnet as shown in the figure. Viewed from above, does the current

flow clockwise or anti-clockwise in the ring?

Ans. According to Lenz‟s law, the direction of the induced current is

opposite to the cause which produces it. So, the side of the ring facing

north pole of magnet must be north pole of the induced magnetic field.

When viewed from above, the current in the ring is clockwise.

Q # 4. What is the direction of the current through resistor R as

shown in the figure? As the switch S is (a) closed (b) open.

Ans. When switch S is closed, then the current in the primary

coil increases from zero to maximum. During this time

interval, magnetic flux through the secondary coil increases

from zero to maximum and induced current produce in it. According to Lenz‟s law, the current

through the secondary should flow in anti-clockwise direction. And current through resistor will be

from left to right.

(b) However, if the switch is opened, the induced current through secondary should flow in clockwise

direction. So the current through resistor R will flow from right to left.

Q # 5. Does the induced emf always act to decrease the magnetic flux through a circuit?

Ans. The induced emf always opposes the cause that produces it.

· If the magnetic flux through the circuit through the circuit is increasing, then induced emf acts

to decrease the magnetic flux.

· If the magnetic flux through the circuit through the circuit is decreasing, then induced emf

acts to increase the magnetic flux.

Hence, the induced emf does not always act to decrease the magnetic flux through the circuit.

Q # 6. When the switch in the circuit is closed, a current is established in the coil and the metal

ring jumps upward. Why? Describe what would happen to the ring if

the battery polarity were reversed?

Ans. When the switch in the circuit is closed, the current is set up in the

coil which establish magnetic field in it.

This result in change of magnetic flux through the metallic ring and hence

an induced emf is produced in it.

The induced magnetic field in the ring opposes the magnetic field

of the coil (according to Lenz‟s law). Therefore the ring experience a

force of repulsion and jumps up.

The same event occurs even if the polarity of the battery is reversed.

Q # 7. Figure shows a coil of wire in the xy-plane with a magnetic field directed along the y-axis.

Around which of the three coordinate axes should the coil be rotated in order to generate an

emf and a current in the coil?

Ans.

· The coil must be rotated along x-axis to get change of

magnetic flux and an induced current through it.

· If the coil is rotated about y-axis, the flux passing

through the coil zero because plane of the coil

remains parallel to magnetic field B all the times.

· If the coil is rotated about z-axis then no change of

magnetic flux takes place through coil.

Hence if the coil is rotated about x-axis, then there is a change of magnetic flux passing through a

coil. So only in this case, an emf is induced in the coil.

Q # 8. How would you position a flat loop of wire in a changing magnetic field so that there is no

emf induced in the loop?

Ans. If the plane of loop of wire is placed parallel to changing magnetic field i.e., , then no flux

through it will change. Hence no emf will be induced through the loop as:

( )

Q # 9. In a certain region, the earth’s magnetic field point vertically down. When a plane flies

due north, which wing tip is positively charged?

Ans. The magnetic force on electrons in the wing is given by:

( )

When the plane flies due north in the earth magnetic field

directed vertically downward, then electrons will experience

force in east direction.

Thus west wingtip of the plane is positively charged.

Q # 10. Show that and

have the same units.

Ans. As we know that:

Ø

———- (1)

Ø

( )( )

( )( )

As ( ) and ( )

———- (2)

Hence from (1) and (2), it is proved that both and

have the same units.

Q # 11. When an electric motor, such as an electric drill, is being used, does it also act as a

generator? If so what is the consequences of this?

Ans. When an electric motor is running, its armature is rotating in a magnetic field. A torque acts on

the armature and at the same time, magnetic flux is changing through the armature which produces an

induced emf. The induced emf opposes the rotation of armature. This means that motor also acts as

generator when it is running.

consequences

· When the motor is just started, back emf is almost zero and hence a large current passes through

the coil.

· As the motor speeds up, the back emf increases and current becomes smaller and smaller.

However, the current is sufficient to provide the torque on the coil drive the load and overcome

losses due to friction.

· If the motor is overloaded, it slows down. Consequently, the back emf decreases and allows motor

to draw more current.

· If the motor is overloaded beyond its limits, the current could be so high that it may burn out the

motor.

Q # 12. Can a DC motor be turned into a DC generator? What changes are required to be

done?

Ans. Yes, a DC motor be turned into a DC generator.

In order to convert DC motor into a DC generator, two changes are to be done:

Ø The magnetic field must be supplied by the permanent magnet and not by

electromagnet.

Ø An arrangement to rotate the coil armature should be provided.

Q # 13. Is it possible to change both the area of the loop and the magnetic field passing through

the loop and still not have an induced emf in the loop?

Ans. If both area of the loop A and magnetic field strength B are changed such that change is

magnetic flux is zero i.e., . Then by Faraday‟s law:

Hence no induced emf in the loop will be produced.

Q # 14. Can an electric motor be used to drive an electric generator with output from the

generator being used to operate the motor?

Ans. No it is not possible. Because if it is possible, it will be a self operating system without getting

energy from some external source and this is against the law of conservation of energy.

Q # 15. A suspended magnet is oscillating freely in a horizontal plane. The oscillations are

strongly damped when a metal plate is placed under the magnet. Explain why this occurs?

Ans. the oscillating magnet produces change of magnetic flux close to it. The metal plate placed

below it experiences the change of magnetic flux. As the result, eddy current are produced inside

metal. According to Lenz‟s law, these eddy current oppose the cause which produce it. So, the

oscillations of magnet are strongly damped.

Q # 16. Four unmarked wires emerge from a transformer. What steps would you take to

determine the turn ratio?

Ans. By checking continuity of the coils, the coils are separated as primary and secondary coils. An

alternating voltage of known value is connected to one coil (primary coil), the output voltage

across the ends of the other coil (secondary coil) is measured. The turn ratio of the coil is determined

by using relation:

Q # 17. (a) Can a step-up transformer increase the power level?

(b) In a transformer, there is no transfer of charge from the primary to the secondary. How is,

then the power transferred?

Ans.

(a). In case of an ideal transformer, the power output is equal to the power input. In actual

transformer, due of dissipation of energy in the coil, the output power is always less than input power.

Therefore, a step-up transformer can‟t increase power level.

(b). The two coils of transformer are magnetically linked i.e., the change of flux through one coil is

Q # 18. When the primary of a transformer is connected to AC mains, the current in it

(a) Is very small if the secondary circuit is open, but

(b) Increases when the secondary circuit is closed. Explain these facts.

Ans. (a). If the secondary circuit is open, then output power will be zero. Because output power is

always slightly smaller than the output power, therefore a very small value of current is being drawn

by a primary coil of transformer form AC mains.

(b). When the secondary circuit is closed, the output power will be increased. As we know that output

power is equal to input power, therefore the transformer will draw large current from the AC mains to

increase the primary power. Hence, greater current is needed in primary to equalize power in

secondary coil.