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# Electromagnetic Induction and Alternating Current: Exercises and Example Solved Numerical problems

Physics : Electromagnetic Induction and Alternating Current: Solved Example Problems with Answers, Solution and Explanation, Numerical problems, Conceptual Questions with answers

## Electromagnetic Induction: Solved Example Problems

EXAMPLE 4.1

A circular antenna of area 3 m2 is installed at a place in Madurai. The plane of the area of antenna is inclined at 47º with the direction of Earth’s magnetic field. If the magnitude of Earth’s field at that place is 40773.9 nT find the magnetic flux linked with the antenna.

Solution

B = 40773.9 nT; θ = 90º  – 47º  = 43°;

A = 3m2

We know that  ΦB = BAcosθ EXAMPLE 4.2

A circular loop of area 5 x 10-2m2 rotates in a uniform magnetic field of 0.2 T. If the loop rotates about its diameter which is perpendicular to the magnetic field as shown in figure. Find the magnetic flux linked with the loop when its plane is (i) normal to the field (ii) inclined 60o to the field and (iii) parallel to the field. Solution

A = 5 ´ 10-2 m2; B = 0.2 T

(i) θ = 0°  ## Faraday’s Experiments on Electromagnetic Induction: Solved Example Problems

EXAMPLE 4.3

A cylindrical bar magnet is kept along the axis of a circular solenoid. If the magnet is rotated about its axis, find out whether an electric current is induced in the coil.

Solution

The magnetic field of a cylindrical magnet is symmetrical about its axis. As the magnet is rotated along the axis of the solenoid, there is no induced current in the solenoid because the flux linked with the solenoid does not change due to the rotation of the magnet.

EXAMPLE 4.4

A closed coil of 40 turns and of area 200 cm2, is rotated in a magnetic field of flux density 2 Wb m-2. It rotates from a position where its plane makes an angle of 30º with the field to a position perpendicular to the field in a time 0.2 sec. Find the magnitude of the emf induced in the coil due to its rotation.

Solution

= 40 turns; B = 2 Wb m-2

= 200 cm2 = 200 ´ 10-4 m2; EXAMPLE 4.5

A straight conducting wire is dropped horizontally from a certain height with its length along east – west direction. Will an emf be induced in it? Justify your answer.

Solution

Yes! An emf will be induced in the wire because it moves perpendicular to the horizontal component of Earth’s magnetic field.

## Fleming’s right hand rule: Solved Example Problems

EXAMPLE 4.6

If the current i flowing in the straight conducting wire as shown in the figure decreases, find out the direction of induced current in the metallic square loop placed near it. Solution

From right hand rule, the magnetic field by the straight wire is directed into the plane of the square loop perpendicularly and its magnetic flux is decreasing. The decrease in flux is opposed by the current induced in the loop by producing a magnetic field in the same direction as the magnetic field of the wire. Again from right hand rule, for this inward magnetic field, the direction of the induced current in the loop is clockwise.

EXAMPLE 4.7

The magnetic flux passes perpendicular to the plane of the circuit and is directed into the paper. If the magnetic flux varies with respect to time as per the following relation: ΦB = (2t3 +3t2 + 8t +5) mWb , what is the magnitude of the induced emf in the loop when t = 3 s? Find out the direction of current through the circuit. Solution

ΦB = (2t 3 +3t2 + 8t +5mWb; N =1; t = 3 s (ii) As time passes, the magnetic flux linked with the loop increases. According to Lenz’s law, the direction of the induced current should be in a way so as to oppose the flux increase. So, the induced current flows in such a way to produce a magnetic field opposite to the given field. This magnetic field is perpendicularly outwards. Therefore, the induced current flows in anti-clockwise direction.

## Motional emf from Faraday’s law and Energy conservation: Solved Example Problems

EXAMPLE 4.8

A conducting rod of length 0.5 m falls freely from the top of a building of height 7.2 m at a place in Chennai where the horizontal component of Earth’s magnetic field is 40378.7 nT. If the length of the rod is perpendicular to Earth’s horizontal magnetic field, find the emf induced across the conductor when the rod is about to touch the ground. [Take g = 10 m s-2]

Solution

l = 0.5 m; h = 7.2 m; u = 0 m s-1;

g = 10 m s-2; BH = 40378.7 nT

The final velocity of the rod is Induced emf when the rod is about to touch the ground, ε = BH lv

= 40, 378.7 ×10−9 × 0.5 ×12

=242 .27 ×10−6V

=242 .27µV

EXAMPLE 4.9

A copper rod of length l rotates about one of its ends with an angular velocity ω in a magnetic field B as shown in the figure. The plane of rotation is perpendicular to the field. Find the emf induced between the two ends of the rod.

Solution Consider a small element of length dx at a distance x from the centre of the circle described by the rod. As this element moves perpendicular to the field with a linear velocity v = xω , the emf developed in the element dx is

d ε = Bvdx = B( xω)dx This rod is made up of many such elements, moving perpendicular to the field. The emf developed across two ends is ## Self-inductance of a long solenoid(Eddy Currents): Solved Example Problems

EXAMPLE 4.10

A solenoid of 500 turns is wound on an iron core of relative permeability 800. The length and radius of the solenoid are 40 cm and 3 cm respectively. Calculate the average emf induced in the solenoid if the current in it changes from 0 to 3 A in 0.4 second.

Solution

N = 500 turns; µr = 800 ;

l = 40 cm = 0.4 m; r = 3 cm = 0.03 m;

di = 3 – 0 = 3 A; dt = 0.4 s

Self inductance, EXAMPLE 4.11

The self-inductance of an air-core solenoid is 4.8 mH. If its core is replaced by iron core, then its self-inductance becomes 1.8 H. Find out the relative permeability of iron.

Solution ## Mutual inductance between two long co-axial solenoids(Eddy Curent): Solved Example Problems

EXAMPLE 4.12

The current flowing in the first coil changes from 2 A to 10 A in 0.4 sec. Find the mutual inductance between two coils if an emf of 60 mV is induced in the second coil. Also determine the induced emf in the second coil if the current in the first coil is changed from 4 A to 16 A in 0.03 sec. Consider only the magnitude of induced emf.

Solution

Case (i):

di1 = 10 – 2 = 8 A; dt = 0.4 s;

ε2 = 60 ×10−3V

Case (ii):

di1 = 16 – 4 = 12 A;

dt = 0.03 s

(i) Mutual inductance of the second coil with respect to the first coil (ii) Induced emf in the second coil due to the rate of change of current in the first coil is EXAMPLE 4.13

Consider two coplanar, co-axial circular coils A and B as shown in figure. The radius of coil A is 20 cm while that of coil B is 2 cm. The number of turns is 200 and 1000 for coils A and B respectively. Calculate the mutual inductance of coil B with respect to coil A. If the current in coil A changes from 2 A to 6 A in 0.04 sec, determine the induced emf in coil B and the rate of change of flux through the coil B at that instant. Solution

NA = 200 turns; NB = 1000 turns;

rA = 20 × 10-2 m; rB = 2 × 10-2 m;

dt = 0.04 s; diA = 6−2 = 4A

Let iA be the current flowing in coil A, then the magnetic field BA at the centre of the circular coil A is The rate of change of magnetic flux of coil is ## Induction of emf by changing the area of the coil: Solved Example Problems

EXAMPLE 4.14

A circular metal of area 0.03 m2 rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω, calculate the induced emf between the axis and the rim and induced current flowing in the disc.

Solution

A = 0.03 m2; B = 0.4 T; f = 20 rps; R = 4 Ω ## Induction of emf by changing relative orientation of the coil with the magnetic field : Solved Example Problems

EXAMPLE 4.15

A rectangular coil of area 70 cm2 having 600 turns rotates about an axis perpendicular to a magnetic field of 0.4 Wb m-2. If the coil completes 500 revolutions in a minute, calculate the instantaneous emf when the plane of the coil is (i) perpendicular to the field (ii) parallel to the field and (iii) inclined at 60o with the field.

Solution

A = 70 ´ 10-4m2; N = 600 turns

B = 0.4 Wbm-2; f = 500 rpm Illustration:

An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case (i):

= 2 MW; R = 40 Ω; V = 10 kV Case (ii):

= 2 MW; R = 40 Ω; V = 100 kV Thus it is clear that when an electric power is transmitted at higher voltage, the power loss is reduced to a large extent.

EXAMPLE 4.16

An ideal transformer has 460 and 40,000 turns in the primary and secondary coils respectively. Find the voltage developed per turn of the secondary if the transformer is connected to a 230 V AC mains. The secondaryisgiventoaloadofresistance104Ω. Calculate the power delivered to the load.

Solution

NP = 460 turns; NS = 40,000 turns

VP = 230 V; RS = 104 Ω EXAMPLE 4.17

An inverter is common electrical device which we use in our homes. When there is no power in our house, inverter gives AC power to run a few electronic appliances like fan or light. An inverter has inbuilt step-up transformer which converts 12 V AC to 240 V AC. The primary coil has 100 turns and the inverter delivers 50 mA to the external circuit. Find the number of turns in the secondary and the primary current.

Solution

Vp = 12 V; Vs = 240 V

Is = 50mA; Np = 100 turns ## RMS value of AC: Solved Example Problems

EXAMPLE 4.18

Write down the equation for a sinusoidal voltage of 50 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.

Solution  EXAMPLE 4.19

The equation for an alternating current is given by i = 77 sin 314t. Find the peak value, frequency, time period and instantaneous value at t = 2 ms.

Solution

i = 77 sin 314t ; t = 2 ms = 2×10-3 s

The general equation of an alternating current is i = Im sin ωt . On comparsion,

(i) Peak value, Im = 77 A

(ii) Frequency, f = ω/2π = 314 / 2 ×3.14 = 50 Hz

Time period, T = 1/f = 150 = 0 .02 s

(iv) At t = 2 m s,

Instantaneous value,

i = 77sin(314×2×10−3 )

i = 45.24 A

## AC circuit containing only an inductor: Solved Example Problems

EXAMPLE 4.20

A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.

Solution

L = 400 x 10-3 H; Ieff = 6 x 10-3A

f = 1000 Hz ## AC circuit containing only a capacitor: Solved Example Problems

EXAMPLE 4.21

A capacitor of capacitance 102/π µF is connected across a 220 V, 50 Hz A.C. mains. Calculate the capacitive reactance, RMS value of current and write down the equations of voltage and current.

Solution ## Quality factor or Q–factor: Solved Example Problems

EXAMPLE 4.22

Find the impedance of a series RLC circuit if the inductive reactance, capacitive reactance and resistance are 184 Ω, 144 Ω and 30 Ω respectively. Also calculate the phase angle between voltage and current.

Solution

XL = 184 Ω; XC = 144 Ω

R = 30 Ω

(i ) The impedance is Impedance, Z = 50 Ω

(ii) Phase angle is φ = 53.1

EXAMPLE 4.23

A 500 μH inductor, 80/π2 pF capacitor and a 628 Ω resistor are connected to form a series RLC circuit. Calculate the resonant frequency and Q-factor of this circuit at resonance.

Solution

L=500×10-6H; C = 80/π2 ×10−12 F; R = 628Ω

(i) Resonant frequency is Q =12.5

EXAMPLE 4.24

Find the instantaneous value of alternating voltage υ = 10 sin(3 π×104 t) volt at i) 0 s ii) 50 μs iii) 75 μs.

Solution

The given equation is υ = 10sin (3 π×104 t) EXAMPLE 4.25

The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.

Solution

L = 40 × 10-3 H; i = 0.1 sin (200t – 40º)

XL = ωL = 200 × 40 × 10-3 = 8 Ω

Vm = Im XL = 0.3 × 8 = 2.4 V

In an inductive circuit, the voltage leads the current by 90o Therefore,

v = Vm sin ( ωt +90º)

v = 2 . 4 sin(200t −40 + 90 º)

v = 2 . 4 sin(200t +50 º)volt

## Power in AC Circuits: Solved Example Problems

EXAMPLE 4.26

A series RLC circuit which resonates at 400 kHz has 80 μH inductor, 2000 pF capacitor and 50 Ω resistor. Calculate (i) Q-factor of the circuit (ii) the new value of capacitance when the value of inductance is doubled and (iii) the new Q-factor.

Solution

L = 80 × 10-6H; C = 2000 × 10-12 F

R = 50 Ω; fr = 400 × 103Hz EXAMPLE 4.27

A capacitor of capacitance 10-4 /π F, an inductor of inductance 2/ π H and a resistor of resistance 100 Ω are connected to form a series RLC circuit. When an AC supply of 220 V, 50 Hz is applied to the circuit, determine (i) the impedance of the circuit (ii) the peak value of current flowing in the circuit (iii) the power factor of the circuit and (iv) the power factor of the circuit at resonance.

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