F.Sc ICS Notes: Physics XII: Chapter 15 Electromagnetic Induction Exercise Short Questions:

FSc ICS Notes: Physics XII: Chapter 15 Electromagnetic Induction Exercise Short Questions: 2nd Year Physics Notes Online Taleem Ilmi Hub

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Question 15.1 Does the induced emf in a circuit depend on the resistance of the circuit? Does the induced current depend on the resistance of the circuit?

Answer 15.1 We know by Faraday's law the induced emf is given by:

ε = -N Δφ/Δt

From above equation it can be seen that induced emf does not depend upon the resistance. The induced current is written as:

I = ε / R

From above equation it can be seen that induced current depends upon the resistance of the circuit..

Question 15.2 A square loop of wire is moving through a uniform magnetic field. The normal to the loop is oriented parallel to magnetic field. Is a emf induced in the loop? Give reasons for your answer?
Answer 15.2 No, the induced emf will not be produced in the loop because the loop is moving in the uniform magnetic field therefore there is not change in magnetic flux. By Faraday's Law the induced emf is written as:

ε = -N Δφ/Δt

as        Δφ/Δt = 0

ε = 0

Question 15.3 A light metallic ring is released from above in to a vertical bar magnet (in the fig). Viewed for above, does the current flow clockwise or anticlockwise in the ring?
Answer 15.3  The emf will be induced in the ring due to change in magnetic flux. The current will flow in such a direction that it will oppose the motion of the ring (Lenz's law). This is possible, only if the magnetic field is produced due to the induced current. In the ring will have its lower side north pole and upper side is south pole. Hence by the right hand rule current should be clockwise.

Question 15.4 What is the direction of the current through resistor R in the fig? When switch S is; (a) closed (b) opened.
Answer 15.4 When the switch is closed the current rises from its zero value to maximum value in the primary coil. Magnetic flux through the secondary coil increases. According to Lenz's Law direction of induced current in the secondary coil should be such that it will oppose the cause, which produce it. Therefore the direction of the induced current is opposite to the direction of current in primary coil. Thus, current through the resistance R is from left to right. When switch is re-opened the magnetic flux through the secondary coil reduces to zero. Therefore by Faraday Law emf induced current in secondary will cause the current in it in same direction as that of the decreasing current in primary coil. Hence the current is from right to left.

Question 15.5 Does the induced emf always act to decrease the magnetic flux through a circuit?
Answer 15.5 If the magnetic flux in the close circuit is decreasing then by Lenz's law the direction of  induced EMF is such that it will try to increase the magnetic flux and vice versa. So it does not always act to decrease the magnetic flux. It will oppose the cause, which produces it.

Question 15.6 When the switch in the circuit is closed a current established in the coil and the metal ring jumps upward (see the fig) Why? Describe what would happen to the ring if battery polarity were reversed?
Answer 15.6 When the switch is closed, the current rises from zero to maximum value instantaneously and change of magnetic flux induces a current in a ring. The induced current, produced in the metallic ring creates its own magnetic field. This magnetic field opposes the magnetic field due to battery current. Thus metal ring is repelled i.e. it will jump upward. When the polarity of the battery is reversed, even then the ring will jump upward.

Question 15.7 The Fig. 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?
Answer 15.7 If the coil is rotated about x-axis then the magnetic flux through the coil is changed, hence EMF will be induced. Whereas, its rotation about y and z axis the magnetic flux remain zero and there is no change in magnetic flux, hence EMF is not induced.

Question 15.8 How would you position a flat loop of wire in a changing magnetic field so that there is no emf induced in the loop?
Answer 15.8 When the plane of the flat loop of wire is held parallel to the direction of magnetic field, the flux will not change and will remain zero.

φ = B•A = BAcos90

φ = 0

Δφ = 0

By using Faraday's Law there will be no induce emf.

Question 15.9 In a certain region the earth’s magnetic field point vertically down. When a plane flies due north, which wingtip is positively charged?
Answer 15.9 Wing of the plane is a metallic and act as a conductor moving through the earth magnetic field. The electrons in the wing are moving with the same velocity as that of plane. Therefore, they will experience force given by:

F = - e (v x B)

The direction of the force by right hand rule is along the direction from A to B. Therefore by looking behind from the plane the left wing tip will be positively charged and right wing will have negative polarity.

Question 15.10 Shoe that ε and ΔΦ / Δt have the same units.
Answer 15.10 Units of EMF  ε is volt if unit of ΔΦ / Δt are measured as follow:

ΔΦ / Δt =  Δ(B•A) / Δt

= NA(-1)m(-1)*m(2) / s

= Nm / As

= Nm / C

ΔΦ / Δt  = J / C .......(i)             

ε = work / charge

ε = J / C   ...........(ii)

Compare (i) and (ii)

ε and ΔΦ / Δt have the same units J / C i.e. volt.

     

Question 15.11 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?
Answer 15.11 Due to back EMF effect in motor the EMF is induced across the ends of the coil when motor is being used. Therefore it will act as a generator.
Consequence:
The current drawn by the motor is given by:

I = (V- ε ) / R

Where ε is the EMF induced across the ends of the coil and V is the voltage of the source. Therefore the motor will draw smaller current when it is running at its normal speed.

Question 15.12 Can a D.C. motor be turned into a D.C. generator? What changes required to be done?
Answer 15.12 As the construction of D.C motor and D.C generator is same. The changes which are to be done are following.

• The current source of the D.C motor is replaced by the load.
• The coil of the motor is to be coupled with a source of mechanical energy i.e shaft of turbine or engine.
• The magnetic field must be supplied by permanent magnet and not by electromagnet.

Question 15.13 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?
Answer 15.13 It is possible to change both area and the magnetic field such that the flux φ remain constant and EMF induced will be zero,

As  φ = B.A

On increasing B the area A of the loop should be decreased and vice versa. Such that the product BA i.e. flux remains constant.

or            φ = B.A = constant

By Faraday's Law the induced emf is given by

ε = -N Δφ/Δt

as           Δφ = 0

ε = 0

                                 

Question 15.14 Can an electric motor be used to drive an electric generator with the output from the generator being used to operate the motor?
Answer 15.14 This system will not work for a long. Since the available mechanical energy initially supplied by the motor decreases in overcoming the frictional losses. A time will reach when no mechanical energy will be available which can be converted into electrical energy, which can drive the motor.

Question 15.15 A suspended magnet is oscillating freely in horizontal plane. Oscillations are strongly damped when a metal plate is placed under the magnet. Explain why this occurs?
Answer 15.15 By placing a metal plate under the vibrating magnet, the magnetic flux through the plate will change. This will induce a current on the surface of plate, known as Eddy current. According to Lenz's Law the direction of the current will oppose the cause, which produces it. So, the oscillations of the magnet will be strongly damped.

Question 15.16 Four unmarked wires emerge from a transformer. What steps would you take to determine the turns ratio?
Answer 15.16 Steps to determine turn ratio in a transformer are as follow:

1. Identifying the ends of primary and secondary coil by using the continuity test be using AVO meter.
2. Measuring the electric resistance of primary coil and secondary coils. If it is step down transfer the coil having greater resistance would be secondary and the other one is primary.
3. By applying known voltage to the primary coil voltage at the secondary coil is measured.
4. Using NS /NP = VS /VP we can find turn ratio that is NS / NP.

Question 15.17 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?
 Answer 15.17  (a) Transformer do not increase power level because at most in case of ideal transformer:

P input = P output

Vp Ip = Vs Is

The transformer increases or decrease alternating voltage. The product VI i.e., power remain constant.

(b) In a transformer power primary and secondary coils are magnetically coupled i.e. the change of magnetic flux is linked with the other coil and hence emf is produced.

Question 15.18 When the primary of a transformer is connected to a.c. 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.
Answer 15.18  When secondary of the transformer is open, no power is delivered to the load. therefore the power drawn by the primary coil from is negligible. Power being defined as:

P = VI

I = P / V

Therefore the current I flowing through primary coil is negligible small. When the secondary circuit is closed the load will consume the power. Therefore a large current will flow through the primary coil.

Written By: Asad Hussain