Unit - 1
Magnetic Field and Magnetic Circuits
Q1) An iron wire ring of 15 cm mean diameter having a cross-section of 90cm2 is wound with 400 turns of wire. Calculate the exciting current required to establish a flux density of 1Wb/m2 if relative permeability of iron is 1000. What is the value of energy stored?
A1) B=
1=4 x x 1000 x 400 x I/0.15
I=0.94A
L=AN2/l=4 x x 1000 x (90x 10-4) x (400)2/0.15=3.84H
E=LI2
=x 3.84 x (0.94)2=1.696J
Q2) A relay has a coil of 900 turns and an air gap of area 10cm2 and length 1.5mm. Calculate the rate of change of stored energy in the air gap of the relay when armature is stationary at 1.5mm from the core and the current is 10mA but increasing at the rate of 20A/s.
A2) As we know L= = = 6.78H
= 20A/s
As armature is stationary = 0
= L I = 6.78 x 10x10-3 x 20=1.36W
Q3) A relay has a coil of 900 turns and an air gap of area 10cm2 and length 1.5mm. Calculate the rate of change of stored energy in the air gap of the relay when current is constant at 15mA but inductance is changing at the rate of 100H/s.
A3) = 100H/s
As current is constant =0
= I2 = x (15x10-3)2 x 100 = 0.011W
Q4) The pole face area of an electromagnet is 0.5m2/pole. It has to lift an iron ingot weighing 860kg. If pole faces are parallel to the surface of the ingot at a distance of 0.8mm determine the coil mmf required.
A4) Force at two poles = 2x B2A/2= B2A/
B2 0.5/ = 860 x 9.8
B= 0.145Wb/m2
H= B/ = 0.145/ = 115.4x103 AT/m
L=2x0.8mm=1.6x10-3m
AT required = 115.4 x103x1.6x10-3 = 184.64
Q5) Reluctance of a magnetic circuit is 105 AT/Wb and excitation coil has 220 turns. Current in the coil is changing at200A/s. Find the inductance of the coil and voltage induced in the coil.
A5) L= N2/S = 2202/105 = 0.48H
eL=L =0.48 x 200 =96.8V
Q6) An iron ring of mean diameter 15cm has a cross sectional area of 100cm2 is wounded with 400turns of wire. Calculate the exciting current required to establish a flux density of 1Wb/m2. If relative permeability is 1000.
A6) B=
1= x 1000x400xI/0.15
I =0.94A
Q7) An iron ring of mean diameter 15cm has a cross sectional area of 100cm2 is wounded with 400turns of wire. Calculate the energy stored. The flux density is 1Wb/m2. If relative permeability is 1000.
A7) E= L I2
B=
1= x 1000x400xI/0.15
I =0.94A
L= = x 1000x 100x10-4x4002/0.15=4.27H
E= L I2
= x 4.27 x 0.94 = 2.01J
Q8) The amount of flux present in around magnetic bar was measured at 0.02 weber. If the material has a diameter of 10cm, calculate the flux density?
A8) Area=r2
Diameter=2r
r=10/2=5cm=0.05m
Area=3.14 x 0.052=0.008m2
Flux Density B==A=0.02/0.008=2.5 Tesla
Q9) Calculate the radius of the material having flux density of 0.7 T and flux present around the magnetic bar is 0.05T?
A9) Flux Density B=A
A= B=0.05/0.7=0.07 m2
Area=r2
r=0.151m
Q10) What are magnetic lines of forces. List their properties?
A10) All magnets have two regions called magnetic poles with the magnetism both in and around a magnetic circuit producing a definite chain of organised and balanced pattern of invisible lines of flux around it. These lines of flux are called as the magnetic field of the magnet. At each end of a magnet is a pole. Magnetic poles are always present in pairs, there is always a region of the magnet called the North-pole and there is always an opposite region called the South-pole. The lines which go to make up a magnetic field showing the direction and intensity are called Lines of Force or Magnetic Flux and represented as Phi (Φ).
Like poles of the magnet repel each other and unlike poles attract each other.
Magnetic flux density =Magnet flux/Area
B=A Tesla
Q11) Explain Ampere’s circuital law?
A11) The law states that the mmf around a closed path is equal to the current enclosed by the path. The law cab be studied for
a) Magnetic field strength of long solenoid:
The magnetic field along the solenoid is H. The ampere turns linked with the path are NI. Then according to Ampere’s Law
H x l=NI
H=NI/l (A/m)
B= μoH
B=μoNI/l (Wb/m2 or tesla)
Q12) Explain Boit Savart Law?
A12) Boit-Savart Law:
The expression for the magnetic field strength dH produced at point P for small length dl of a conductor carrying current of I amperes is given as
dH= A/m
dB0= μo Wb/m2
Q13) List the applications of Boit Savart Law?
A13)
1) We can use Biot–Savart law to calculate magnetic responses even at the atomic or molecular level.
2) It is also used in aerodynamic theory to calculate the velocity induced by vortex lines.
3) Biot-Savart law is similar to the Coulomb’s law in electrostatics.
4) The law is applicable for very small conductors too which carry current.
5) The law is applicable for symmetrical current distribution.
Q14) Determine the magnetic field at the centre of a semi-circular piece of wire with radius 0.3m. The current carried by the semi-circular piece of wire is 120A?
A14) As we already know from Biot-Savart Law
dB0= μo Wb/m2
Integrating both sides of above equation
B0 = = = (4 x 10-7x120)/ (4 x 0.3)
B0 = 1.257 x 10-4 Tesla
Q15) Define as well as derive magnetic flux?
A15) The number of magnetic lines of forces set up in a magnetic circuit is called Magnetic Flux. It is analogous to electric current, I in an electric circuit. Its SI unit is Weber (Wb) and its CGS unit is Maxwell. It is denoted by φB.
ΦB=B.S
ΦB=B.S Cos
B – the magnitude of the magnetic field
S – area of surface
θ – angle between the magnetic field lines and perpendicular distance normal to the surface area
Magnetic flux for a closed surface
ΦB==0
Magnetic flux for open surface is
E=
E=
E – electromotive force
v – velocity of the boundary
E – electric field
B – magnetic field
øB - magnetic flux through the open surface
The magnetic flux through a closed surface is always zero, but in the open surface, it is not zero.