Electromagnetic Theory Formulas for GATE EE Exam

# Electromagnetic Theory Formulas for GATE EE Exam | GATE Notes & Videos for Electrical Engineering - Electrical Engineering (EE) PDF Download

``` Page 3

Magnetic dipoles:
Acurrentloopcreatesamagneticdipole~ µ=I
~
Awhere
I is the current in the loop and
~
A is a vector normal to
the plane of the loop and equal to the area of the loop.
The torque on a magnetic dipole in a magnetic ?eld is
~ t =~ µ×
~
B
Biot-Savart Law:
The magnetic ?eld d
~
B produced at point P by a dif-
ferential segment d
~
l carrying current I is
d
~
B =
µ
0
4p
Id
~
l×ˆ r
r
2
where ˆ r points from the segment d
~
l to the point P.
Magnetic ?eld produced by a moving charge:
Similarly, the magnetic ?eld produced at a point P by
a moving charge is
~
B =
µ
0
4p
q~ v×ˆ r
r
2
Amp` ere’s Law: (without displacement current)
I
~
B·d
~
l =µ
0
I
encl
The EMF produced in a closed loop depends on the
change of the magnetic ?ux through the loop
E =-
dF
B
dt
Self Inductance:
A changing current i in any circuit generates a chang-
ing magnetic ?eld that induces an EMF in the circuit:
E =-L
di
dt
where L is the self inductance of the circuit
L=N
F
B
i
For example, for a solenoid of N turns, length l, area A,
Amp` ere’s law gives B = µ
0
(N/l)i, so the ?ux is F
B
=
µ
0
(N/l)iA, and so
L=µ
0
N
2
l
A
When an EMF is produced by a changing magnetic ?ux
thereisaninduced,nonconservative,electric?eldE
~
such
that
I
~
E·dl
~
=-
d
dt
Z
A
~
B·d
~
A
Mutual Inductance:
Whenachangingcurrenti
1
incircuit1causesachang-
ingmagnetic?uxincircuit2,andvice-versa,theinduced
EMF in the circuits is
E
2
=-M
di
1
dt
and E
1
=-M
di
2
dt
where M is the mutual inductance of the two loops
M =
N
2
F
B2
i
1
=
N
1
F
B1
i
1
where N
i
is the number of loops in circuit i.
Capacitance:
A capacitor is any pair of conductors separated by an
insulatingmaterial. Whentheconductorshaveequaland
opposite charges Q and the potential di?erence between
the two conductors is V
ab
, then the de?nition of the ca-
pacitance of the two conductors is
C =
Q
V
ab
The energy stored in the electric ?eld is
U =
1
2
CV
2
If the capacitor is made from parallel plates of area A
separated by a distance d, where the size of the plates is
much greater than d, then the capacitance is given by
C =²
0
A/d
Capacitors in series:
1
C
eq
=
1
C
1
+
1
C
2
+...
Capacitors in parallel:
C
eq
=C
1
+C
2
+...
If a dielectric material is inserted, then the capacitance
increases by a factor of K where K is the dielectric con-
stant of the material
C =KC
0
Page 4

Magnetic dipoles:
Acurrentloopcreatesamagneticdipole~ µ=I
~
Awhere
I is the current in the loop and
~
A is a vector normal to
the plane of the loop and equal to the area of the loop.
The torque on a magnetic dipole in a magnetic ?eld is
~ t =~ µ×
~
B
Biot-Savart Law:
The magnetic ?eld d
~
B produced at point P by a dif-
ferential segment d
~
l carrying current I is
d
~
B =
µ
0
4p
Id
~
l×ˆ r
r
2
where ˆ r points from the segment d
~
l to the point P.
Magnetic ?eld produced by a moving charge:
Similarly, the magnetic ?eld produced at a point P by
a moving charge is
~
B =
µ
0
4p
q~ v×ˆ r
r
2
Amp` ere’s Law: (without displacement current)
I
~
B·d
~
l =µ
0
I
encl
The EMF produced in a closed loop depends on the
change of the magnetic ?ux through the loop
E =-
dF
B
dt
Self Inductance:
A changing current i in any circuit generates a chang-
ing magnetic ?eld that induces an EMF in the circuit:
E =-L
di
dt
where L is the self inductance of the circuit
L=N
F
B
i
For example, for a solenoid of N turns, length l, area A,
Amp` ere’s law gives B = µ
0
(N/l)i, so the ?ux is F
B
=
µ
0
(N/l)iA, and so
L=µ
0
N
2
l
A
When an EMF is produced by a changing magnetic ?ux
thereisaninduced,nonconservative,electric?eldE
~
such
that
I
~
E·dl
~
=-
d
dt
Z
A
~
B·d
~
A
Mutual Inductance:
Whenachangingcurrenti
1
incircuit1causesachang-
ingmagnetic?uxincircuit2,andvice-versa,theinduced
EMF in the circuits is
E
2
=-M
di
1
dt
and E
1
=-M
di
2
dt
where M is the mutual inductance of the two loops
M =
N
2
F
B2
i
1
=
N
1
F
B1
i
1
where N
i
is the number of loops in circuit i.
Capacitance:
A capacitor is any pair of conductors separated by an
insulatingmaterial. Whentheconductorshaveequaland
opposite charges Q and the potential di?erence between
the two conductors is V
ab
, then the de?nition of the ca-
pacitance of the two conductors is
C =
Q
V
ab
The energy stored in the electric ?eld is
U =
1
2
CV
2
If the capacitor is made from parallel plates of area A
separated by a distance d, where the size of the plates is
much greater than d, then the capacitance is given by
C =²
0
A/d
Capacitors in series:
1
C
eq
=
1
C
1
+
1
C
2
+...
Capacitors in parallel:
C
eq
=C
1
+C
2
+...
If a dielectric material is inserted, then the capacitance
increases by a factor of K where K is the dielectric con-
stant of the material
C =KC
0
Current:
When current ?ows in a conductor, we de?ne the cur-
rent as the rate at which charge passes:
I =
dQ
dt
Wede?nethecurrentdensityasthecurrentperunitarea,
and can relate it to the drift velocity of charge carriers
by
~
J =nq~ v
d
where n is the number density of charges and q is the
charge of one charge carrier.
Ohm’s Law and Resistance:
Ohm’sLawstatesthatacurrentdensityJ inamaterial
isproportionaltotheelectric?eldE. Theratio?=E/J
is called the resistivity of the material. For a conductor
with cylindrical cross section, with area A and length L,
the resistance R of the conductor is
R =
?L
A
A current I ?owing through the resistor R produces a
potential di?erence V given by
V =IR
Resistors in series:
R
eq
=R
1
+R
2
+...
Resistors in parallel:
1
R
eq
=
1
R
1
+
1
R
2
+...
Power:
The power transferred to a component in a circuit by
a current I is
P =VI
whereV isthepotentialdi?erenceacrossthecomponent.
Kirchho?’s rules:
The algebraic sum of the currents into any junction
must be zero:
X
I =0
The algebraic sum of the potential di?erences around
any loop must be zero.
X
V =0
Force on a charge:
An electric ?eld E
~
exerts a forceF
~
on a chargeq given
by:
F
~
=q
~
E
Coulomb’s law:
A point charge q located at the coordinate origin gives
rise to an electric ?eld E
~
given by
~
E =
q
4p²
0
r
2
r ˆ
where r is the distance from the origin (spherical coor-
dinate), r ˆis the spherical unit vector, and ²
0
is the per-
mittivity of free space:
²
0
=8.8542×10
-12
C
2
/(N·m
2
)
Superposition:
The principle of superposition of electric ?elds states
that the electric ?eld E
~
of any combination of charges
is the vector sum of the ?elds caused by the individual
charges
~
E =
X
i
~
E
i
To calculate the electric ?eld caused by a continuous dis-
tribution of charge, divide the distribution into small el-
ements and integrate all these elements:
~
E = dE
~
=
Z Z
q
dq
4p²
0
r
2
r ˆ
Electric ?ux:
Electric ?ux is a measure of the “?ow” of electric ?eld
through a surface. It is equal to the product of the
area element and the perpendicular component of E
~
in-
tegrated over a surface:
F
E
= EcosfdA=
~
E·n ˆdA=
Z Z Z
~
E·d
~
A
where f is the angle from the electric ?eld E
~
to the sur-
face normal ˆ n.
Gauss’ Law:
Gauss’ law states that the total electric ?ux through
any closed surface is determined by the charge enclosed
by that surface:
F
E
=
I
~
E·dA
~
=
Q
encl
²
0
Page 5

Magnetic dipoles:
Acurrentloopcreatesamagneticdipole~ µ=I
~
Awhere
I is the current in the loop and
~
A is a vector normal to
the plane of the loop and equal to the area of the loop.
The torque on a magnetic dipole in a magnetic ?eld is
~ t =~ µ×
~
B
Biot-Savart Law:
The magnetic ?eld d
~
B produced at point P by a dif-
ferential segment d
~
l carrying current I is
d
~
B =
µ
0
4p
Id
~
l×ˆ r
r
2
where ˆ r points from the segment d
~
l to the point P.
Magnetic ?eld produced by a moving charge:
Similarly, the magnetic ?eld produced at a point P by
a moving charge is
~
B =
µ
0
4p
q~ v×ˆ r
r
2
Amp` ere’s Law: (without displacement current)
I
~
B·d
~
l =µ
0
I
encl
The EMF produced in a closed loop depends on the
change of the magnetic ?ux through the loop
E =-
dF
B
dt
Self Inductance:
A changing current i in any circuit generates a chang-
ing magnetic ?eld that induces an EMF in the circuit:
E =-L
di
dt
where L is the self inductance of the circuit
L=N
F
B
i
For example, for a solenoid of N turns, length l, area A,
Amp` ere’s law gives B = µ
0
(N/l)i, so the ?ux is F
B
=
µ
0
(N/l)iA, and so
L=µ
0
N
2
l
A
When an EMF is produced by a changing magnetic ?ux
thereisaninduced,nonconservative,electric?eldE
~
such
that
I
~
E·dl
~
=-
d
dt
Z
A
~
B·d
~
A
Mutual Inductance:
Whenachangingcurrenti
1
incircuit1causesachang-
ingmagnetic?uxincircuit2,andvice-versa,theinduced
EMF in the circuits is
E
2
=-M
di
1
dt
and E
1
=-M
di
2
dt
where M is the mutual inductance of the two loops
M =
N
2
F
B2
i
1
=
N
1
F
B1
i
1
where N
i
is the number of loops in circuit i.
Capacitance:
A capacitor is any pair of conductors separated by an
insulatingmaterial. Whentheconductorshaveequaland
opposite charges Q and the potential di?erence between
the two conductors is V
ab
, then the de?nition of the ca-
pacitance of the two conductors is
C =
Q
V
ab
The energy stored in the electric ?eld is
U =
1
2
CV
2
If the capacitor is made from parallel plates of area A
separated by a distance d, where the size of the plates is
much greater than d, then the capacitance is given by
C =²
0
A/d
Capacitors in series:
1
C
eq
=
1
C
1
+
1
C
2
+...
Capacitors in parallel:
C
eq
=C
1
+C
2
+...
If a dielectric material is inserted, then the capacitance
increases by a factor of K where K is the dielectric con-
stant of the material
C =KC
0
Current:
When current ?ows in a conductor, we de?ne the cur-
rent as the rate at which charge passes:
I =
dQ
dt
Wede?nethecurrentdensityasthecurrentperunitarea,
and can relate it to the drift velocity of charge carriers
by
~
J =nq~ v
d
where n is the number density of charges and q is the
charge of one charge carrier.
Ohm’s Law and Resistance:
Ohm’sLawstatesthatacurrentdensityJ inamaterial
isproportionaltotheelectric?eldE. Theratio?=E/J
is called the resistivity of the material. For a conductor
with cylindrical cross section, with area A and length L,
the resistance R of the conductor is
R =
?L
A
A current I ?owing through the resistor R produces a
potential di?erence V given by
V =IR
Resistors in series:
R
eq
=R
1
+R
2
+...
Resistors in parallel:
1
R
eq
=
1
R
1
+
1
R
2
+...
Power:
The power transferred to a component in a circuit by
a current I is
P =VI
whereV isthepotentialdi?erenceacrossthecomponent.
Kirchho?’s rules:
The algebraic sum of the currents into any junction
must be zero:
X
I =0
The algebraic sum of the potential di?erences around
any loop must be zero.
X
V =0
Force on a charge:
An electric ?eld E
~
exerts a forceF
~
on a chargeq given
by:
F
~
=q
~
E
Coulomb’s law:
A point charge q located at the coordinate origin gives
rise to an electric ?eld E
~
given by
~
E =
q
4p²
0
r
2
r ˆ
where r is the distance from the origin (spherical coor-
dinate), r ˆis the spherical unit vector, and ²
0
is the per-
mittivity of free space:
²
0
=8.8542×10
-12
C
2
/(N·m
2
)
Superposition:
The principle of superposition of electric ?elds states
that the electric ?eld E
~
of any combination of charges
is the vector sum of the ?elds caused by the individual
charges
~
E =
X
i
~
E
i
To calculate the electric ?eld caused by a continuous dis-
tribution of charge, divide the distribution into small el-
ements and integrate all these elements:
~
E = dE
~
=
Z Z
q
dq
4p²
0
r
2
r ˆ
Electric ?ux:
Electric ?ux is a measure of the “?ow” of electric ?eld
through a surface. It is equal to the product of the
area element and the perpendicular component of E
~
in-
tegrated over a surface:
F
E
= EcosfdA=
~
E·n ˆdA=
Z Z Z
~
E·d
~
A
where f is the angle from the electric ?eld E
~
to the sur-
face normal ˆ n.
Gauss’ Law:
Gauss’ law states that the total electric ?ux through
any closed surface is determined by the charge enclosed
by that surface:
F
E
=
I
~
E·dA
~
=
Q
encl
²
0
```

## GATE Notes & Videos for Electrical Engineering

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## FAQs on Electromagnetic Theory Formulas for GATE EE Exam - GATE Notes & Videos for Electrical Engineering - Electrical Engineering (EE)

 1. What are some important formulas in electromagnetic theory for the GATE EE exam in electrical engineering?
Ans. Some important formulas in electromagnetic theory for the GATE EE exam include: 1. Gauss's Law: ∮E⃗ · dA⃗ = 1/ε₀ ∮ρ⃗ · dA⃗ 2. Ampere's Law: ∮B⃗ · dl⃗ = μ₀(I + ε₀ d(∮E⃗ · dA⃗)/dt) 3. Faraday's Law: ∮E⃗ · dl⃗ = -d(∮B⃗ · dA⃗)/dt 4. Coulomb's Law: F⃗ = k(q₁q₂/r²) ̂r 5. Ohm's Law: V = IR, where V is voltage, I is current, and R is resistance.
 2. How can I use Gauss's Law to solve problems in electromagnetic theory for the GATE EE exam?
Ans. Gauss's Law is a powerful tool for solving problems in electromagnetic theory. It allows you to calculate the electric field produced by a charge distribution by integrating the electric flux over a closed surface surrounding the charge. To use Gauss's Law, follow these steps: 1. Identify the symmetry of the charge distribution and choose an appropriate closed surface to apply Gauss's Law. 2. Calculate the flux of the electric field through the closed surface, which is given by the dot product of the electric field vector and the differential area vector. 3. Evaluate the integral to solve for the electric field using the charge enclosed by the surface and the permittivity of free space. 4. Repeat the process for different charge distributions and surfaces, if necessary. Remember to consider the direction and orientation of the electric field vector when applying Gauss's Law.
 3. What is Ampere's Law and how can it be used in electromagnetic theory for the GATE EE exam?
Ans. Ampere's Law relates the magnetic field around a closed loop to the current passing through the loop. It can be used to find the magnetic field produced by a current-carrying wire or a current distribution. To use Ampere's Law, follow these steps: 1. Identify the symmetry of the current distribution and choose a closed loop that encloses the current. 2. Calculate the line integral of the magnetic field along the closed loop, which is given by the dot product of the magnetic field vector and the differential length vector. 3. Evaluate the integral to solve for the magnetic field using the current passing through the loop, the permittivity of free space, and the rate of change of electric flux through the loop. 4. Repeat the process for different current distributions and loops, if necessary. Ampere's Law is particularly useful in problems involving long, straight wires or cylindrical symmetry.
 4. How does Faraday's Law relate to electromagnetic theory for the GATE EE exam?
Ans. Faraday's Law states that a changing magnetic field induces an electromotive force (EMF) or voltage in a closed loop. It is one of the fundamental principles of electromagnetic theory. Faraday's Law can be used to calculate the induced voltage or EMF in various situations, such as when a magnetic field changes with time or when a conducting loop is moving in a magnetic field. To use Faraday's Law, follow these steps: 1. Identify the magnetic field or flux that is changing with time or the loop that is moving in a magnetic field. 2. Calculate the line integral of the electric field along the closed loop, which is given by the dot product of the electric field vector and the differential length vector. 3. Evaluate the integral to solve for the induced voltage or EMF using the rate of change of magnetic flux through the loop. 4. Repeat the process for different magnetic field configurations or moving loops, if necessary. Faraday's Law is essential for understanding electromagnetic induction and the operation of devices such as transformers and generators.
 5. How can I apply Ohm's Law in electromagnetic theory for the GATE EE exam?
Ans. Ohm's Law relates the voltage across a conductor to the current flowing through it and the resistance of the conductor. It can be applied in various electromagnetic theory problems involving circuits and electrical systems. To apply Ohm's Law, follow these steps: 1. Identify the circuit or conductor for which you want to calculate the voltage or current. 2. Measure or determine the resistance of the conductor. 3. Determine the current flowing through the conductor, or vice versa. 4. Use Ohm's Law, V = IR, to calculate the voltage or current. When applying Ohm's Law, ensure that the units of voltage, current, and resistance are consistent. Ohm's Law is particularly useful in analyzing circuits and solving problems related to electrical power, energy dissipation, and circuit behavior.

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