Download UNIT 6: MAGNETISM

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A group of
phenomena
associated with
magnetic field.
UNIT 6: MAGNETISM
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6.1 Magnetic Field
Definition – is defined as the region around a magnet where
a magnetic force can be experienced.
{
A stationary electric charge is surrounded by an electric field only.
{
When an electric charge moves, it is surrounded by an electric field
and a magnetic field. The motion of the electric charge produces
the magnetic field.
field
{
Magnetic field has two poles, called north (N) and south (S).
(S) This
magnetic poles are always found in pairs whereas a single magnetic
pole has never been found.
Like poles (Nz
(N-N or SS-S) repel each other.
Opposite poles (Nz
(N-S) attract each other.
6.1.1 Magnetic field lines
{
Magnetic field lines are used to represent a magnetic field.
{
By convention, magnetic field lines leave the north pole and enters
the south pole of a magnet.
z
Magnetic field lines can be represented by straight lines or curves.
The tangent to a curved field line at a point indicates the direction
of the magnetic field at that point as shown in figure 6.1.a.
{
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Fig. 6.1a
P
direction of magnetic
field at point P.
2
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z
{
Magnetic field can be represented by crosses of by dotted circles
as shown in figures 6.1b and 6.1c.
X
X
X X
X
X
X X
X X X X
Fig. 6.1c : magnetic field lines
Fig. 6.1b : magnetic field lines
leave the page perpendicularly
enter the page perpendicularly
A uniform field is represented by parallel lines of force. This means that
the number of lines passing perpendicularly through unit area at all
cross-sections in a magnetic field are the same as shown in figure
6.1d.
Fig. 6.1d
{
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unit crosscross-sectional area
A non-uniform field is represented by non-parallel lines. The number of
magnetic field lines varies at different unit cross-sections as shown in
figure 6.1e.
3
Fig. 6.1e
A1
A2
weaker field in A2
stronger field in A1
The number of lines per unit crosscross-sectional area is proportional
to the magnitude of the magnetic field.
field
{
Magnetic field lines do not intersect one another.
6.1.2 Magnetic field lines Pattern
{
The pattern of the magnetic field lines can be determined by using two
methods.
Using compass needles (shown in figure 6.1f)
z
{
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Fig. 6.1f : plotting a magnetic
field line of a bar magnet.
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z
{
Using sprinkling iron filings on paper (shown in figure 6.1g).
Fig. 6.1g : Thin iron filing indicate the magnetic
field lines around a bar magnet.
Figure below shows the various pattern of magnetic field lines around
the magnets.
a. Bar magnet
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b. Horseshoe or U magnet
c. Two bar magnets (unlike pole) - attractive
d. Two bar magnets (like poles) - repulsive
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Neutral point (point
(point where
the resultant magnetic 6
force is zero).
zero).
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6.1.3 Magnetization of a Soft Iron
{
There are two methods to magnetized the soft iron.
Using the permanent magnet.
z
{
One permanent magnet
z A permanent magnet is bring near to the soft iron and
touching the surface of the soft iron by following the path in
the figure 6.1h.
N
z
z
S
Fig. 6.1h
This method is called induced magnetization.
magnetization
The arrows in the soft iron represent the magnetization
direction with the arrowhead being the north pole and
arrow tail being the south pole. It is also known as
domains ( the tiny magnetized region because of spin
magnetic moment of the electron).
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In an unmagnetized piece of soft iron, these domains are
arranged randomly but it is aligned in one direction when
the soft iron becomes magnetized.
z The soft iron becomes a temporary magnet with its south
pole facing the north pole of the permanent magnet and
vise versa as shown in figure 6.1h.
Two permanent magnets
z Bring and touch the first magnet to one end of the soft iron
and another end with the second magnet as shown in
figure 6.1i and 6.1j.
z
{
Fig. 6.1i
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Fig. 6.1j
N
S
S
NN
S
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z
Using Electrical circuit.
{
A soft iron is placed inside a solenoid (a long coil of wire
consisting of many loops of wire) that is connected to the
power supply as shown in figure 6.1k.
N
S
I
I
Current anticlockwise
Switch, S
Current - clockwise
Fig. 6.1k
{
{
When the switch S is closed, the current I flows in the solenoid
and produces magnetic field.
The directions of the fields associated with the solenoid can be
found by viewing the current flows in the solenoid from
both end as shown if figure 6.1k or applying the right hand
grip rule below.
Important
Thumb – north pole
Other fingers – direction of current in solenoid.
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{
Other examples:
a.
S
I
b.
S
I
{
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N
I
N
I
Note :
If you drop a permanent magnet on the floor or strike it with a
hammer, you may jar the domains into randomness.
randomness The magnet
can thus lose some or all of its magnetism.
Heating a magnet too can cause a loss of magnetism.
z
z
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The permanent magnet also can be demagnetized by placing it
inside a solenoid that connected to an alternating source.
source
Example 1 : (exercise)
Sketch the magnetic field lines pattern around the bar magnets as
shown in figures below.
a.
b.
z
{
6.2 Magnetic Flux Density and Magnetic Flux
6.2.1 Magnetic flux density, B
{
Definition – is defined as the magnetic flux per unit area across an
area at right angles to the magnetic field.
field
Mathematically,
Φ where
B=
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B
Φ B : magnetic flux
A⊥ A⊥ : area at right angles to the magnetic field
{
It also known as magnetic induction (magnetic
magnetic field intensity)
intensity 11
{
It is a vector quantity and its direction follows the direction of the
magnetic field.
Its unit is weber per metre squared (Wb m-2) or tesla (T).
Unit conversion :
-2
4
{
{
1 T = 1 Wb m
= 10 gauss(G )
6.2.2 Magnetic flux, ΦB
{
Magnetic flux of a uniform magnetic field,
field
Definition – is defined as the scalar product between the magnetic
flux density, B with the vector of the surface area, A.
r r
Mathematically,
Φ B = B • A = BA cos θ
where
Φ B : magnetic flux
θ : angle between the direction of B and A.
{
{
It is a scalar quantity and its unit is weber (Wb).
Consider a uniform magnetic field B passing through a surface area A
as shown in figures 6.2a and 6.2b.
area, A
From the fig. 6.2a, θ =0°, thus
r
A
Φ B = BA cos 0 o
Φ B = BA
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Fig. 6.2a
r
B
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r
A
area, A
{
r
B
θ
{
Note that the direction of vector
A is always perpendicular
(normal) to the surface area, A.
Fig. 6.2b
The magnetic flux is proportional to the number of field lines
passing through the area.
Let us consider an element of area dA on an arbitrarily shaped surface
r
as shown in figure 6.2c.
{ If the magnetic field at this element is B
, ther magnetic
r flux through the element
r
is B • dA
dA
θ
r
Where dA is a vector that is
perpendicular to the surface and has a
r
B
magnitude equal to the area dA.
{
Fig. 6.2c
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{
between B and A is θ, thus
Φ B = BA cos θ
{
{
From the fig. 6.2b, the angle
Therefore, the total magnetic flux
through the surface is given by
r r
Φ B = ∫ B • dA = ∫ BdA cos θ
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Example 2 :
30 o
r
B
A
Figure above shows a flat surface with an area of 3.0 cm2 is placed in
the uniform magnetic field. The plane surface makes an angle 30 ° with
the direction of the magnetic field. If the magnetic flux through the
surface is 0.90 mWb, calculate the magnitude of the magnetic field.
Solution:
30 o
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A=3.0x10-4 m2, ΦB=0.90x10-3 Wb
r
From the figure, the angle between B
A r
o
o
o
B
and A is θ = 90 − 30 = 60
θ
30 o
By applying the equation of magnetic
flux for uniform B hence
Φ B = BA cos θ
ΦB
B=
A cos θ
B = 6.0 T
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6.3 Biot-Savart Law
{
The law can be stated as :
The magnitude of magnetic flux density dB at a point P which is a
distance r from a very short length dl of a conductor carrying a
current I is given by
{
dB ∝
Idl sin θ
r2
(6.3a)
where θ is the angle between the short length and the line joining
it to point P.
This law can be summarized by using the diagram shown in figure 6.3a.
I
θ
r
dl
P
r
r̂
{
I
Fig. 6.3a
r
dB into the paper
The direction of dB is given by the
right hand grip rule (figure 6.3b).
Fig. 6.3b
Important
Thumb – direction of current
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Other
{
fingers – direction of magnetic field.
If the medium around the conductor is vacuum or air then the equation
6.3a can be written as
dB =
{
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µ0 Idl sin θ
4π r 2
(6.3b) Magnitude form
In vector form, using the unit vector r̂, we have
dB =
µ0 Idl × rˆ
4π r 2
(6.3c) Vector form
where Idl : current element
{
µ0 : permeability of free space
= 4 π x 10 −7 T m A−1
To find the total magnetic field B at any point in space due to the currentr
in a complete circuit, eq. 6.3b needs to be integrated over all segment dl
that carry current, symbolically
B=
µ0 Idl sin θ
4π ∫ r 2
(6.3d) Magnitude form
or
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B=
µ0 Idl × rˆ
4π ∫ r 2
(6.3e) Vector form
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6.4 Ampere’s Law
{
{
Ampere’s law is an alternative to the Biot-Savart law.
For conductors in a vacuum or air the law states
r r
B
∫ • dl = µ0 I enc
where
∫ Bdl cosθ = µ I
or
0 enc
integral of the flux density around
∫ Bdl cosθ : line
a closed path
It is the sum of the terms Bdl
{
(6.4a)
cos θ for every
very short length dl of the closed path.
B : magnitude of the magnetic flux density
I enc : current enclosed by the path r
r
θ : angle between the direction of B and dl
This law can be summarized by using the diagram shown in figure 6.4a.
r
B
Part of closed path
r
dl
θ
clockwise
or
anticlockwise
Fig. 6.4a
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{
Note :
z Both Biot-Savart and Ampere’s law can be used to determine the
magnetic field of a long straight conductor, a circular coil and a long
thin solenoid.
6.5 Magnetic field produced by the electrical
current
When a current flows in a conductor wire or coil, the magnetic field will
be produced.
{
The direction of magnetic field around the wire or coil can be
determined by using the right hand grip rule as shown in figure 6.3b.
6.5.1 Magnetic field of a long straight conductor (wire) carrying current
{
The magnetic field lines pattern around a straight conductor carrying
current is shown in figures 6.5a and 6.5b.
{
I
r
B
or
r
B
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I
Fig. 6.5a
I
r
B
r
B
Current out of the page
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r
B
I
r
B
r
B
I
Fig. 6.5b
a
X Current into the page
z
Consider a straight conductor with length 2a carrying a current I as
shown in figure 6.5c.
z
To find the field dB at point P caused by
the element of the conductor of length
2
2
r
dl θ
r= x +y
y π −θ
P r
X dB
0
x
dl=dy shown in figure 6.5c, firstly we use
Biot-Savart law.
dB =
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-a
r
B
The equation of magnetic flux density at any point from a long straight
wire carrying current can be determined by using Biot-Savart law.
{
2a
XI
or
I
µ0 Idy sin θ
4π
r2
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Fig. 6.5c
z
From the figure 6.5c,
r = x2 + y2
and
sin θ = sin(π − θ ) =
x
x2 + y2
Therefore dB at point P produced by element dl is given by

x
dy
2
2

µ0 I  x + y
dB =
4π
x2 + y2
µI
xdy
dB = 0
4π x 2 + y 2 3 / 2
(
z
(6.5a)
To find the magnitude of the total magnetic flux density B at point P
supplied by the whole conductor, integrate eq. 6.5a from -a to a then
µ0 I
−a
B=
xdy
a
∫ dB = 4π ∫ (x
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)




2
+ y2
µ0 I
2a
4π x x 2 + a 2
)
3/ 2
(6.5b)
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z
When the length of the conductor is infinitely long then a is much
x 2 + a 2 ≈ a.
larger than x so that
z
Therefore the B at point P produced by whole conductor is
where
B=
µ0 I
2πx
x : distance of any point from the conductor
6.5.2 Magnetic field of a Circular Shaped Coil
The magnetic field lines pattern around a circular shaped coil carrying
current is shown in figures 6.5d.
{
I
R
N
or
S
{
X
I
S
I
I
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N
I
Fig. 6.5d
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The equation of magnetic flux density at any point from a circular
shaped coil carrying current can be determined by using Biot-Savart law.
z Consider a circular shaped conductor with radius R that carries a
current I as shown in figure 6.5e.
y r
dl
R
φ
r = x2 + R2
O
x
I
z I
I
Fig. 6.5e
r
dB
dB y
φ
P
dBx
x
z
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To find the field dB at a point P on the axis of the circular (loop) at
distance x from the centre O, we need to apply Biot-Savart law.
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z
From the figure 6.5e, dl and r are perpendicular and the direction of
z
the field dB caused by the element dl lies in the yz-plane.
Since r = x 2 + R 2, the magnitude dB of the field due to element
z
dl is given by
µ Idl sin θ
o
and θ = 90
dB = 0
2
4π r
µI
dl
dB = 0
2
4 π (x + R 2 )
The components of the vector dB are
x-component :
dBx = dB cos φ
and
cos φ =
(x
R
+ R2
2
dBx =
µ0 I
dl
R
2
2
4 π (x + R ) (x 2 + R 2 )1 / 2
dBx =
µ0 IR
dl
4π x 2 + R 2
(
)
1/ 2
)
3/ 2
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y-component :
dB y = dB sin φ
and
sin φ =
y
z
I
X
I
Fig. 6.5f
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+ R2
)
1/ 2
elements around the loop, the resultant component is zero.
zero It
because the current in any element on the one side of the loop
sets up a dBy that cancels the dBy set up by the current through
the element diametrically opposite it.
it (figure 6.5f)
Therefore the resultant field at point P must along the x axis and
is given by
r
dB
P
O
2
µ0 Ix
dl
2
4 π (x + R 2 )3 / 2
By symmetry, when the component dBy are summed over all
dB y =
z
(x
x
r
dB
∫ dB = ∫
x
x
Bx =
Bx =
µ0 IR
dl
4π x 2 + R 2
µ0 IR
(
(
4π x 2 + R 2
(
µ0 IR 2
2 x2 + R2
)
3/ 2
dl and ∫ dl = 2πR
) ∫
Circumference of
3/ 2
the circular coil
)
3/ 2
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z
In general ,
B=
µ0 IR 2
(
2 x +R
2
On the axis of
a circular coil
)
2 3/ 2
or
B=
µ0 NIR 2
(
2 x +R
2
On the axis of N
circular coils
)
2 3/ 2
The magnitude of magnetic field B at point O (centre
centre of the
circular coil or loop)
loop , x=0 is given by
z
B=
where
(
µ0 IR 2
20+R
)
2 3/ 2
B=
µ0 NI
2R
At the centre of N
circular coils
R : radius of the circular coil
N : number of coils (loops)
µ0 : permeability of free space
= 4 π x 10 −7 T m A−1
I : current
x : distance between a point with a centre of the coil
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6.5.3 Magnetic field of a Solenoid
{
The magnetic field lines pattern around a solenoid carrying current is
shown in figures 6.5g.
N
S
I
I
or
Fig. 6.5g
XI
XI
XI
XI
N
S
I
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I
I
I
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{
The equation of magnetic flux density around a solenoid carrying
current I can be determined by using Ampere’s law.
z Consider a very long solenoid with closely packed coils, the field is
nearly uniform and parallel to the solenoid axes within the entire
cross section, as shown in figure 6.5h.
Fig. 6.5h
r
r
c B dl
r
r
dl
dl
b
r l
d
r
dl
a
B(outside) = 0(very small )
z
z
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To find the magnetic field inside the solenoid and at the centre, we
choose and draw the rectangle closed path abcd as shown in figure
6.5h (clockwise) for applying Ampere’s law.
By considering this path consists of four segment : ab, bc, cd and
27
da, then Ampere’s law becomes
r r
B
∫ • dl = µ0 I enc and I enc = NI
r cr r d r r ar r
b r
B
•
d
l
+ ∫ B • dl + ∫ B • dl + ∫ B • dl = µ0 NI
∫
a
b
c
d
: number of coils (loops) our path encircles
c
d
a
o
o
o
∫ Bdl cos 90 + ∫ Bdl cos 0 + ∫ Bdl cos 90 = µ0 NI
b
where N
d
c
B ∫ dl = µ0 NI
c
B=
µ0 NI
l
and
B = µ0 nI
z
c
dl = l
N
=n
l
At the centre/ midmidpoint/ inside of N
turn solenoid
where n : number of coils per unit length
The magnetic flux density at the end of the solenoid is given by
B=
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∫
and
or
d
d
1
( µ0 nI )
2
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{
Example 3 :
Two long straight wires are placed parallel to each other and carrying
the same current I. Sketch the magnetic field lines pattern around both
wires
a. when the currents are in the same direction.
b. when the currents are in opposite direction.
Solution:
I
I
a.
I
I
or
I
I
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b.
I
I
I
I
or
I
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XI
30
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{
Example 4 :
A long wire (X) carrying a current of 30 A is placed parallel to and 3.0
cm away from a similar wire (Y) carrying a current of 6.0 A.
a. Find the magnitude and direction of the magnetic flux density midway
between the wires :
i. when the current are in the same direction.
ii. when they are in opposite direction.
b. When the currents are in the same direction there is a point
somewhere between X and Y at which the magnetic flux density is
zero. How far from X is this point ?
(Given µ0 = 4π x 10-7 H m-1)
Solution: IX=30
a. i.
A, IY=6.0 A, d=3.0x10-2 m
d r
BX
rX A rY
or
r
BY
IX
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IY
rX
IX
r
BX
A
rY
IY
r
B
d Y
rX = rY = = 1.5 x10 − 2 m
31
2
By using the equation of magnetic field at any point near the
straight wire, then at point A
µI
Magnitude of BX : B = 0 X
B = 4.0 x10 −4 T
2πrX
µI
BY = 0 Y
2πrY
X
Magnitude of BY :
X
Direction : into the page
(upwards)
BY = 8.0 x10 −5 T
Direction : out of page
(downwards)
Therefore the total
r magnetic
r flux
r density at point A is
Sign convention of B
BA = BX + BY
Into the page ⇒ + (positive)
Out of page ⇒ - (negative)
a.ii.
BA = BX − BY
BA = 3.2 x10 −4 T
BX
r
BY
d
rX
r
B
r X
BY
A
Direction : into the page
(upwards)
r
rY
or
IX
rX
rX = rY =
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IX
IY
A
rY
X
IY
d
= 1.5 x10 − 2 m
2
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By using the equation of magnetic field at any point near the
straight wire, then at point A
µI
Magnitude of BX : B X = 0 X
B = 4.0 x10 −4 T
2πrX
Magnitude of BY :
BY =
X
Direction : into the page
(upwards)
µ0 I Y
2πrY
BY = 8.0 x10 −5 T
Direction : into the page
(upwards)
Therefore the total
r magnetic
r flux
r density at point A is
BA = BX + BY
BA = BX + BY
BA = 4.8 x10 −4 T
d
b.
r
BX
r
BX
A r
Y
rX
or
r
BY
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Direction : into the page
(upwards)
IX
IY
IX
rX
A
rX = r
ry = d − r
rY
r
BY
IY
33
By using the equation of magnetic field at any point near the
straight wire, then at point A
µI
Magnitude of BX : B = 0 X Direction : into the page
X
(upwards)
2πr
Magnitude of BY :
BY =
µ0 I Y
2π(d − r)
Direction : out of page
(downwards)
Since the total magnetic
at point A is zero, hence
r
r flux density
r
BA = BX + BY
0 = BX − BY
µ0 I X
µ0 I Y
=
2πr 2π(d − r)
dI X
r=
I X + IY
r = 2.5 x10 −2 m
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{
Example 5 :
a. A 2000 turns solenoid of length 40 cm and resistance 16 Ω is
connected to a 20 V supply. Find the magnetic flux density at the mid
point of the axis of the solenoid.
b. A solenoid 1.30 m long and 2.60 cm in diameter carries a current of
18.0 A. The magnetic field inside the solenoid is 23.0 mT. Find the
length of the wire forming the solenoid. (Halliday,Resnick&Walker,p.708,no.42)
(Given µ0=4 π x 10-7 H m-1)
Solution:
a. Given N=2000 turns, l=40x10-2 m, R=16 Ω, V=20 V
By applying the equation of magnetic flux density at the centre of the
solenoid, thus
V
µ0 NI
and I =
R
l
µ NV
B= 0
lR
B = 7.9 x10 −3 T
B=
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b. Given l=1.30 m, d=2.60x10-2 m, I=18.0 A, B=23.0x10-3 T
By applying the equation of magnetic flux density inside the solenoid,
thus
µ0 NI
B=
l
Bl
N=
µ0 I
N = 1322 turns
Since the shaped of each coil forms the solenoid is circle, then the
circumference of one coil is
circumference = πd
circumference = 8.17 x10 −2 m
Therefore the length of the wire forming the solenoid, L is
{
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L = N (circumference)
L = 108 m
Example 6 :
A closely wound circular coil a diameter of 4.00 cm has 600 turns and
carries a current of 0.500 A. Determine the magnitude of the magnetic
field
a. at the centre of the coil.
b. at a point on the axis of the coil 8.00 cm from its centre.
(Given µ0=4 π x 10-7 H m-1) (Young&Freedman,p.1098,no.28.28)
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Solution: N=600 turns, d=4.00x10-2 m, I=0.500 A
a. By using the equation of magnetic flux density at the centre of the
coil, thus
µ NI
d
and R =
B= 0
2R
µ NI
B= 0
d
B = 9.43 x10 −3 T
b. Given x=8.00x10-2 m.
2
By applying the equation of magnetic flux density at a point of
distance x on the axis of the circular coil from its centre, thus
B=
B=
(
µ0 NIR 2
2 x +R
2
)
2 3/ 2
and
R=
d
2
µ0 NId 2
3/2

d2 
8 x 2 + 
4 

B = 1.35 x10 −4 T
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{
37
Example 7 :
The segment of wire in figures a and b carry a current of I=5.00
where the radius of the circular arc is R=3.00
A,
cm.
I
R
O
I
O
R
Fig. b : exercise
Fig. a
For each figure, find the magnetic flux density at origin (point O)
(Given µ0=4 π x 10-7 H m-1)
Solution: I=5.00 A, R=3.00x10-2 m
Figure a : Sections (1) and (3) are straight line and the angle
between dl and the line joining dl to point O is 0°or 180°.
Ir 1
Hence B1 and B3 at point O is
dl
2
R
O
SF027
µ0 Idl sin 0 o
µ0 Idl sin 180 o
B1 =
and B3 =
4π ∫
r2
4π ∫
r2
3
B1 = B3 = 0
r
dl
38
SF017
Section (2) is a quarter of circular coil hence the B2 at point O is
 1  µ NI
Use right hand grip
and N = 1
B2 =   0
rule
 4  2R
µ0 I
−5
B2 = 2.62 x10 T Direction : into the page
B2 =
8R
Therefore the total magnetic flux density at point O is
BO = B1 + B2 + B3
BO = 2.62 x10 −5 T Direction : into the page
Figure b : Ans. :5.24 x 10-5 T into the page.
Example 8 :
Four long, parallel power wires each carry 100 A current. A cross
sectional diagram for this wires is a square, 20.0 cm on each side. For
each case in figures a, b and c,
{
X
X
X
X
Fig. a
X
X
X
Fig. b :exercise
X
Fig. c : exercise
SF027
39
i. sketch the magnetic field lines pattern on the diagram.
ii. calculate the magnetic flux density at the centre of the square.
(Given µ0=4 π x 10-7 H m-1)
Solution: I1=I2=I3=I4=I=100
Figure (a) :
i.
X
X
I2
l
X
I3
l2 + l2
r1 = r2 = r3 = r4 = r =
2
r = 0.141 m
Since r1=r2=r3=r4 and I1=I2=I3=I4 then the magnitude of B1,B2, B3
and B4 at point C are the same and given by
µI
B1 = B2 = B3 = B4 = B = 0
2πr
40
B = 1.41x10 −4 T
X
SF027
X
A, l=20.0x10-2 m
I1
l
X
ii.
r
r1 B2 rr2
r
B
B1 Cr 3
r4 B4 r3
X
I4
X
SF017
From the diagram, B1 cancel the B3 and B2 cancel the B4, thus
r
r r
r r
BC = B1 + B2 + B3 + B4
BC = 0
Figure (b) : Ans. : 0
Figure (c) : Ans. : 4.0 x 10-4 T direction : to the left (180°)
6.6 Magnetic Force on a Moving Charge
A stationary electric charge in a magnetic field will not experience any
{
force.
force But if the charge is moving with a velocity, v in a magnetic field,
B then it will experience a force. This force known as magnetic force.
force
The magnitude of the magnetic force can be calculated by using the
equation below :
F = Bqv sin θ
(6.6a)
{
In vector form,
form
where
SF027
{
(
r
r r
F = q v×B
)
(6.6b)
F : magnetic force
B : magnetic flux density
v : velocity of a charge
q : magnitude of the charge
r
r
θ : angle between v and B
41
The direction of the magnetic force can be determined by using the
Fleming’s hand rule.
Fleming’
z
Fleming’s right hand rule : - for negative charge shown in figures
6.6a and 6.6b
Fleming’
z
Fleming’s left hand rule : - for positive charge
r
B
r
F
r
F
r
B
r
v
r
v
Fig. 6.6a
Thumb – direction of Force
Important
Fig. 6.6b
First finger – direction of Field
{
Second finger – direction of Velocity.
Velocity.
Example 9 :
r
Determine the direction of the magnetic force, F exerted on a charge in
each problems below.
r
a.
b.
v
r
+
SF027
r
B
B
−
r
v
42
SF017
c. X
X
X
X
r
BX
X
d.
X
X
rX
v
X
−r
v
X
I
−
X
e.
+
r
v
I
Solution:
a. By using Fleming’s left hand rule, b. By using Fleming’s right hand rule,
thus
r
thus
r
F (into the
r
B
r page)
v
v
+
r
F
r
B
−
(to the left)
d. Using right hand grip rule to
c. By using Fleming’s right hand rule, determine the direction of magnetic
field forms by the current I on the
thus
charge position. Then apply the
X
X
X X
r
Fleming’s right hand rule, thus
r
F X X −X X
X
X
X X B
(to the left)
r
BX
SF027
X
r
v
r
F
X
(to the left) I
X
X
−r
v
X
X
43
e. Using right hand grip rule to determine the direction of magnetic field
forms by the current I on the
r charge position. Then apply the Fleming’s
F (upwards)
left hand rule, thus
r
X XB
X
Xr
X
X
+
I
{
X X
Example 10 : (exercise)
Determine the sign of a charge in each problems below.
r
r
r
a.
b.
B
F
{
SF027
v
B
r
v
r
F
r
v
Ans. : positive charge, positive charge
Example 11 : (exercise)
Determine the direction of the magnetic force exerted on a positive
charge in each figures below when a switch S is closed.
44
SF017
a.
b.
+
Switch, S
{
r
v
r
v
+
Switch, S
Ans. :a. into the page, b. out of page
Example 12 :
Calculate the magnitude of the force on a proton travelling 3.0
x107 m s-1 in the uniform magnetic flux density of 1.5 Wb m-2, if :
a. the velocity of the proton is perpendicular to the magnetic field.
b. the velocity of the proton makes an angle 50° with the magnetic field.
(Given the charge of the proton is +1.60 x 10-19 C)
Solution: v=3x107 m s-1, B=1.5 T, q=1.60x10-19 C
a. Given θ = 90° then by applying the equation of magnetic force on a
moving charge, thus
F = Bqv sin θ
F = 7.2 x10 −12 N
SF027
45
b. Given θ = 50° then by applying the equation of magnetic force on a
moving charge, thus
F = Bqv sin θ
F = 5.5 x10 −12 N
{
{
SF027
Example 13 : (exercise)
An electron experiences the greatest force as it travels 2.9 x 106 m s-1 in
a magnetic field when it is moving north. The force is upward and of
magnitude 7.2 x 10-13 N. Find the magnitude and direction of the
magnetic field. (Giancolli, pg.705, no.22)
(Given the charge of the electron is -1.60 x 10-19 C)
Ans : 1.6 T to the east.
Example 14 : (exercise)
An electron is moving in a magnetic field. At a particular instant, the
speed of the electron is 3.0 x 106 m s-1. The magnitude of the magnetic
field on the electron is 5.0 x 10-13 N and the angle between the velocity
of the electron and the magnetic force is 30°. Calculate the magnitude
of the magnetic flux density on the electron in the field.
(Given the charge of the electron is -1.60 x 10-19 C)
Ans : 1.2 T
46
SF017
6.7 Magnetic Force on a current-carrying
conductor
{
{
When a current-carrying conductor is placed in a magnetic field B, thus
a magnetic force will acts on that conductor.
The magnitude of the magnetic force exerts on the current-carrying
conductor is given by
(6.7a)
F = BIL sin θ
In vector form,
form
where
r
F
{
(
)
r
r r
(6.7b)
F = I L× B
F : magnetic force
B : magnitude of the magnetic flux density
I : current
L : length of the conductor
r
θ : angle between direction of I and B
The direction of the magnetic force can be determined by using the
Fleming’
’s left hand rule as shown in figure 6.7a.
Fleming
r
B
Important
Thumb – direction of Force
I
First finger – direction of Field
Second finger – direction of Current.
Current.
SF027
47
Fig. 6.7a
{
Note :
z It is clear from eq. (6.7a),
{
the magnetic force on the conductor has its maximum value
when the conductor (and therefore the current) and the
magnetic field are perpendicular (at right angles) to each other
then θ=90° (shown in figure 6.7b).
{
the magnetic force on the conductor is zero when the
conductor (and therefore the current) is parallel to the magnetic
field then θ=0° (shown
r in figure 6.7c).
r
B
I
θ = 90 o
Fmax = BIL sin 90 o
Fmax = BIL
Fig. 6.7b
{
SF027
B
I
θ = 0o
F = BIL sin 0 o
F =0
Fig. 6.7c
One tesla is defined as the magnetic flux density of a field
in which a force of 1 newton acts on a 1 metre length of a
conductor which carrying a current of 1 ampere and is
perpendicular to the field.
48
SF017
{
Example 15 :
Determine the direction of the magnetic force, exerted on a conductor
carrying current, I in each problems below.
a.
b.
X
X
X
X
r
BX
X
X
X
X
X
X
X
X
X
X
X
r
BX
X
I
X
X
X
X
X
X
X
X
X
X (to the right)
X
X
I
Solution:
For both problems, use Fleming’s left hand rule :
b.
a.
r
F
X
X
X
X
X
X
X
X
X
X
X
X
X
X
r
BX
X
(to therleft)
BX
X
I
SF027
I
r
F
49
{
Example 16 :
A wire of 20 cm long is placed perpendicular to the magnetic field of
0.40 Wb m-2.
a. Calculate the magnitude of the force on the wire when a current 12 A
is flowing.
b. For the same current in (a), determine the magnitude of the force on
the wire when its length is extended to 30 cm.
c. If the force on the 20 cm wire above is 60 x 10-2 N and the current
flows is 12 A, find the magnitude of magnetic field was supplied.
Solution: L=20x10-2
a. Given I
m, B=0.40 T, θ=90°
= 12 A.
By applying the equation of magnetic force on a current-carrying
conductor, thus F = BIL sin θ
F = 0.96 N
b. Given I = 12 A and L = 30x10-2 m
By applying the equation of magnetic force on a current-carrying
conductor, thus
F = BIL sin θ
F = 1.4 N
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50
SF017
c. Given I
= 12 A, L = 20x10-2 m , θ=90° and F = 60x10-2 N
By applying the equation of magnetic force on a current-carrying
conductor, thus F = BIL sin θ
B=
F
IL sin θ
B = 0.25 T
6.8 Forces between two current-carrying
conductors
{
r
B2
SF027
Fig. 6.8a
{
Consider two identical straight conductors X and Y carrying currents I1
and I2 with length L are placed parallel to each other as shown in figure
6.8a.
{
The conductors are in vacuum and
X
Y
their separation is d.
I1
{
The magnitude of the magnetic flux
r
I2
r
B1
density, B1 at point P on conductor Y
F12
due
to the current in conductor X is
P
given by
µI
r
B1 = 0 1 Direction : into the
F
21
Q
2πd page/paper
{
I1
d
I2
Conductor Y carries a current I2 and in
the magnetic field B1 then conductor Y
51
will experiences a magnetic force, F12.
The magnitude of F12 is given by
{
F12 = B1 I 2 L sin θ and θ = 90 o
µ I 
Use Fleming’
Fleming’s left
F12 =  0 1  I 2 L sin 90 o
hand
rule
π
2
d


µ I 
F12 =  0 1  I 2 L Direction : to the left (towards X)
 2πd 
The magnitude of F21 is given by
F21 = B2 I 1 L sin θ and θ = 90 o
µ I 
F12 =  0 2  I 1 L sin 90 o
 2πd 
µ I 
F12 =  0 2  I 1 L Direction : to the right (towards Y)
 2πd 
{
Conclusion :
r
r
µII L
F12 = F21 = F = 0 1 2
2πd
(6.8a)
The properties of this force : Attractive force
SF027
52
SF017
If the direction of current in conductor Y is change to upside down as
shown in figure 6.8b.
{
X
Y
I1
r
F21
I1
Fig. 6.8b
{
Q
I2 r
B1
P
r
B2
d
The magnitude of F12 and F21 can
be determined by using eq. 6.8a and
its direction by applying Fleming’s left
r hand rule.
F12 Conclusion :
{
The properties of this force :
Repulsive force
I2
Note :
The currents are in the same direction – 2 conductors attract each
other.
z
The currents are in opposite direction – 2 conductors repel each
other.
z
SF027
53
{
Example 17 :
Two very long parallel wires are placed 2.0 cm apart in air. Both wires
carry a current of 8.0 A and 10 A respectively. Find
a. the magnitude of the magnetic force in newton, on each metre length
of wire.
b. the magnetic flux density at point P, midway between the wires if the
currents (exercise)
i. in the same direction.
ii. In opposite direction.
(Given µ0=4 π x 10-7 H m-1)
Solution: I1=8.0
A, I2=10 A, d=2.0x10-2 m
a. Given L = 1.0 m
By applying the equation of force for two parallel current-carrying
µ0 I 1 I 2 L
conductors, thus
F=
2πd
F = 8.0 x10 −4 N
b. Ans. : 0.4 x 10-4 T out of page (downwards), 3.6 x10-4 T into the page
(upwards)
Hint : Example 4a.
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54
SF017
{
Example 18 : (exercise)
Two long, straight, parallel wires in a vacuum are 0.25 m apart.
i. The wires each carry a current of 2.40 A in the same direction.
Calculate the force between the wires per metre of their length. Draw
a sketch showing clearly the direction of the force on each wire.
ii. The current in one of the wires is reduced to 0.64 A. Calculate the
current needed in the second wire to maintain the same force
between the wires per metre of their length as in (i).
(Given µ0=4 π x 10-7 H m-1)
Ans. : 4.6 µN, 9.0 A
6.9 Definition of Ampere and Ampere Balance
6.9.1 Definition of Ampere
{
From the eq. (6.8a), if two long, straight, parallel conductors , 1.0 m
apart in vacuum carry equal 1.0 A currents hence the force per unit
length that each conductor exerts on the other is
F µ0 I 1 I 2
=
L
2 πd
SF027
F ( 4πx10 −7 )( 1.0 )( 1.0 )
=
L
2π( 1.0 )
F
= 2.0 x10 −7 N m −1
L
55
The ampere is defined as the constant current that, when it is
flowing in each of two infinitely long, straight, parallel conductors
conductors
which have negligible of cross sectional areas and are 1.0 metre
apart in vacuum, would produce a force per unit length
between the conductors of 2.0 x 10-7 N m-1.
6.9.2 Ampere (current) Balance
{
An instrument used to measure a current absolutely, on the basis of the
definition of the ampere (due to the forces between two long, straight,
parallel conductors).
{
Figure 6.9a shows a schematic diagram for a current balance where the
current can be determined by measuring the force between two
conductors carrying the same current.
E
D
d
P
C
Q
l Fr F
A
B
{
m
G
r
r
W = mg
SF027
Fig. 6.9a
H
I
A
56
SF017
{
{
The plane ABCD and light rod PQ are adjusted so that the plane is
initially horizontal.
Conductors BC and EF are placed parallel to each other, equal in length
l and separated by a distance d.
{
{
When a current I flows, conductors BC and EF repels, hence the plane
ABCD is unbalanced.
The mass m needed to restore the plane ABCD in torque equilibrium
where Torque due to the mass = Torque due to the force on BC
and l AB = l PQ
WlPQ = Fl AB
µIIl
mg = 0 1 2
2πd
µ0 I 2 l
mg =
2πd
then
I=
where
SF027
and
I1 = I 2 = I
2πmgd
µ0 l
g : gravitional acceleration (9 .81 m s - 2 )
57
6.10 Motion of a Charged Particle in a
Uniform Magnetic Field
6.10.1 A charged particle moves perpendicular to the magnetic field.
{
Consider a charged particle moving in a uniform magnetic field with its
velocity perpendicular to the magnetic field.
{
As the particle enters the region, it will experiences a magnetic force
which the force is perpendicular to the velocity of the particle. Hence the
direction of its velocity changes but the magnetic force remains
perpendicular to the velocity.
{
+
This magnetic force, FB makes the path of the particle is a circular as
shown in figures 6.10a, 6.10b, 6.10c and 6.10d.
r
v
X
X
X
X
X
r
F
X B
X FB
X
SF027
Xr
X
r
+
X
X
X
Fig. 6.10a
v
X
+
+
r
v
r+
FB
r
v
r
FB
rX
v
X
+
r
v
Fig. 6.10b
58
SF017
−
r
v
X
X
X
X
X
X
r
FB −X
X
X
X
X
r
X
{
r
v
X
r−
FB
r
v
v
−
X
X
X
Fig. 6.10c
r
v
r
FB
−
r
FB −
r
v
Fig. 6.10d
Since the path is circle therefore the magnetic force FB contributes the
centripetal force Fc (net force) in this motion. Thus
FB = Fc
mv 2
o
and θ = 90
Bqv sin θ =
r
mv where
r=
m : mass of the charged particle
Bq v : magnitude of the velocity
r : radius of the circular path
q : magnitude of the charged particle
SF027
59
{
The period of the circular motion, T makes by the particle is given by
v = rω
T=
and
2πr
v
2π
T
mv
r=
Bq
ω=
and
or
T=
{
Since
T=
1
f
2πm
Bq
thus the frequency of the circular motion makes by the
particle is
f =
Bq
2πm
6.10.2 A charged particle moves not perpendicular to the magnetic field.
{
If the direction of the initial velocity is not perpendicular to the uniform
magnetic field, the velocity component parallel to the field is constant
because there is no force parallel to the field.
{
Therefore the particle moves in a helix path as shown in figure 6.10e.
SF027
60
SF017
y
r
v⊥
r r r
v = v⊥ + v//
r
rwhere
vr⊥ : velocity component perpendicular to
r the B
v// : velocity component parallel to the B
r
v
r
v//
r
B
z
{
{
Fig. 6.10e
x
mv⊥
The radius of the helix, r is given by r =
Bq
Example 19 :
An electron at point A in figure below has a speed v0 of 1.41 x 106 m s-1.
v0
A
−
10.0 cm
B
Given e = 1.60 x10-19 C
me= 9.11 x 10-31 kg
Find
a. the magnitude and direction of the
magnetic field that will cause the electron
to follow the semicircular path from A to B.
b. the time required for the electron to move
from A to B. (Young&Freedman,p.1055,no.27.15)
SF027
61
Solution: v0=1.41x106 m s-1, d=10.0x10-2 m
a. Since the path makes by the electron is semicircular thus the
magnitude of the magnetic field is given by
d
me v0
and r =
Be
2
2me v0
B = 1.61x10 −4 T
B=
ed
r=
The direction of B : electron (use Fleming’s right hand rule)
v0
A
−
r
F
B
Into the page.
b. Since the path is semicircular then the time required for the electron
moves from A to B is half of the period and given by
SF027
d
2π
1
and r =
t = T where v0 = rω; ω =
2
2
T
πd
t=
2v0
t = 1.11x10 −7 s
T=
πd
v0
62
SF017
6.11 Lorentz Force and Determination of q/m
6.11.1 Lorentz Force
{
Definition – is defined as the total force acting on a charge q moving
with a velocity v in the presence of both an electric field
{
E and a magnetic field B.
Its formula is given by
r r
r
r
r
r
r r
F = FE + FB where FE = qE and FB = q v × B
r
r r r
r
F = q( E + v × B) where F : Lorentz force
(
)
It also known as electromagnetic force.
force
6.11.2 Determination of q/m
{
The value of q/m is constant for any charged particle.
{
{
Consider a positive charged particle with mass m, charge q and speed
v enters a region of space where the electric and magnetic fields are
perpendicular to the particle’s velocity and to each other as shown in
figure 6.11a.
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63
+
+ ++ + + ++ ++Fr++ + + ++ ++ + r
B
X
X
XB
X
X
X
r
r
X
X
X
X
vX
v X
+
X
X
X
X
X
X
+
r
v
r
X
X
X r X
E X
−−
−−
−−
−
− −− −− −− − −F−−
E
X
Fig. 6.11a
{
The charged particle will experiences the electric force FE is downwards
{
magnitude Bqv (fig. 6.11a).
If the particle travels in a straight line with constant velocity hence the
electric and magnetic
r r forces
r are equal
r in magnitude. Therefore
with magnitude qE and the magnetic force FB is upwards with
F = FE + FB
Bqv = qE
E
v=
B
{
SF027
and
F =0
(6.11a)
Only particles with speeds equal to E/B can pass through without being
deflected by the fields. Eq. (6.11a) also works for electron or other
64
negative charged particles.
SF017
{
If no electric field in fig. 6.11a, then the particle will makes a circular path
where
therefore
{
FB = FC
mv 2
Bqv =
r
q
v
=
m rB
q
E
= 2
m rB
and
v=
E
B
(6.11b)
In the J.J. Thomson’s experiment , the speed v of electron with mass m
is determined by an accelerating potential difference V between two
plates where Kinetic energy of the
= Electric potential
electron
energy
1 2
mv = eV
2
2eV
v=
m
(6.11c)
Therefore by substituting eq. (6.11c) into eq. (6.11a) then the value of
e/m is given by
SF027
65
2eV  E 
= 
m
B
2
e
E2
=
m 2VB 2
{
(6.11d)
From the experiment the value of e/m is 1.758820174 x 1011 C kg-1.
Example 20 :
An electron with kinetic energy of 5.0 keV passes perpendicular through
a uniform magnetic field of 0.40 x 10-3 T. It is found to follow a circular
path. Calculate
a. the radius of the circular path.
b. the time required for the electron to complete one revolution.
(Given 1 eV=1.60x10-19 J, e/m =1.76x1011 C kg-1, me =9.11x10-31 kg)
Solution: B=0.40x10-3 T, K=5.0x103(1.60x10-19)=8.0x10-16
a. The speed of the electron is
1 2
mv
2
2K
v=
m
K=
SF027
v = 4.2 x107 m s −1
66
J
SF017
Since the path made by the electron is circular, thus
FB = FC
mv 2
o
Bev sin θ =
and θ = 90
r
v
r=
e
 B
m
r = 0.60 m
b. The time required for the electron to complete one revolution is
2π
v = rω and ω =
T
2πr
T=
T = 9.0 x10 −8 s
v
{
Example 21: (exercise)
A proton moving in a circular path perpendicular to a constant magnetic
field takes 1.00 µs to complete one revolution. Determine the magnitude
of the magnetic field. (Serway & Jewett,pg.921,no.32)
(Given mp =1.67x10-27 kg, charge of the proton, q=1.60 x 10-19 C)
Ans. : 6.56 x 10-2 T
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67
6.12 Torque on a Coil in a Magnetic Field
{
Consider a rectangular coil (loop) of wire with side lengths a and b that
it can turn about axis PQ. The coil is in a magnetic field of flux density B
and the plane of the coil makes an angle θ with the direction of the
magnetic field. A current I is flowing round the coil as shown in figure
r
6.12a.
r
r
F
B
B
I
I
r r
A B
P
φ θ
r
F1
r
B
b
a
I
r
F1
r
B
Q
r
F
I
Fig. 6.12a
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r
B
rotation
r
F1
{
r
B
r
F1
r
B
r
A
θ
φ
φ
b
2
Q
φ
b
sin φ
2
rotation
b
sin φ
2 Fig. 6.12b : Side view
From the figure 6.12a, the force F1 on the right side of the coil is to the
right and B is perpendicular to the current thus the magnitude of the
force F1 is given by
{
{
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{
{
F1 = BIL sin 90 o and L = a
F1 = IaB
(6.12a)
Another force F1 with the same magnitude but opposite direction acts
on the opposite side of the coil. (fig. 6.12a)
The forces F act on the side length b where the lines of action of both
forces lies along the PQ. (fig. 6.12a)
69
The total force on the coil is zero but the net torque is not zero
because the forces F1 are not lie along the same line thus the rotation
of the coil about an axis PQ is clockwise (fig. 6.12b).
The magnitude of the net torque about the axis PQ ( fig. 6.12b) is given
by

b

b
τ = − F1  sin φ  − F1  sin φ 
2

2

b

τ = −2 F1  sin φ  and F1 = IaB
2

b

τ = −2(IaB ) sin φ 
2

τ = − IabB sin φ and ab = A(area of coil)
τ = IAB sin φ
o
and φ = 90 − θ
(6.12b)
or
τ = IAB cos θ
τ : torque on the coil
B : magnetic flux density
I : current flows in the coil
where
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70
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{
For a coil of N turns, the torque is given by
τ = NIAB sin φ
(6.12c)
or
τ = NIAB cos θ
r
r
φ : angle between vector area A and B
r
θ : angle between the plane of the coil and B
Note :
N : number of turns (coils)
z The torque is zero when θ=90° or φ=0° but the magnetic flux is
maximum as shown in figure 6.12c.
r
r
A
τ = NIAB sin 0 o or τ = NIAB cos 90 o
B
θ = 90 o τ = 0
o
but ΦB = BA cos 0
o
φ=0
maximum
ΦB = BA
where
{
Fig. 6.12c
The torque is maximum when θ=0° or φ=90° but the magnetic
flux is zero as shown in figure 6.12d.
z
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71
τ = NIAB sin 90 or τ = NIAB cos 0
maximum
τ = NIAB
o
but ΦB = BA cos 90
ΦB = 0
o
z
If
r
r
µ = NIA
r
B
o
φ = 90
r
A
o
θ = 0o
Fig. 6.12d
then eq. (6.12c) can be written as
τ = µB sin φ
Magnitude form
In vector form is
r r r
τ = µ× B
where µ is called magnetic moment or electromagnetic moment.
moment
Magnetic moment is a vector quantity.
Its direction can be determined by using right hand grip rule.
E.g.
Important
I
{
{
I
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r
µ
Thumb – direction of magnetic moment
Other fingers – direction of current in the
72
coil
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z
In a radial field,
field the plane of the coil is always parallel to the
magnetic field for any orientation of the coil about the vertical axis
as shown in figure 6.12e.
Fig. 6.12e : Plan view of moving coil meter
N
radial field fixed soft
iron cylinder
S
θ = 0o
or
φ = 90 o
coil
Hence the torque on the coil in a radial field is always constant
and maximum given by
τ = NIAB sin 90 oor τ = NIAB cos 0 o
τ = NIAB
maximum
Radial field is used in moving coil galvanometer.
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6.13 Moving-Coil Galvanometer and Direct
Current (DC) Motor
6.13.1 Moving-Coil Galvanometer
{
A galvanometer consists of a coil of wire suspended in the magnetic
field of a permanent magnet. The coil is rectangular and consists of
many turns of fine wire as shown in figure 6.13a.
{
When the current I flows through
the coil, the magnetic field exerts
a torque on the coil as given by
τ = NIAB
{
This torque is opposed by a
spring which exerts a torque, τs
given by
where
τ s = kθ
k : torsional constant
θ : rotation angle of the coil
in radian
Fig. 6.13a
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{
The coil and pointer will rotate
only to the point where the spring
torque balances the torque due to
74
magnetic field, thus
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τ = τs
NIAB = kθ
kθ
I=
NAB
6.13.2 Direct-Current (DC) Motor
{
A motor is an instrument that converts electrical energy to mechanical
energy.
{
A motor works on the same principle as a galvanometer, except that
there is no spring so the coil can rotate continuously in one direction.
{
When a current flows in the coil, a torque is produced, which causes the
coil PQRS to rotate as shown in figure 6.13b.
N
R
I
S
Fig. 6.13b
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{
{
{
{
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I
B1
Q
I
C1, C2 : Commutators
B1, B2 : Brushes
P
C1 C2
B2
S
I
PQRS : Rectangular coil
75
The commutators also rotate with the coil PQRS but the brushes are
stationary with the circuit.
When the coil rotates half revolution (180°), each commutator changes
its connection to the other brush where C1 touching B2 and C2 touching
B1. This arrangement will cause the direction of the current through the
coil to be reversed after every half revolution and ensures that the
direction of the torque is always the same. Therefore the coil can turn
continuously.
The torque on the coil PQRS of the motor is given by
τ = NIAB sin φ or τ = NIAB cos θ
Example 22:
A 20 turns rectangular coil with sides 6.0 cm x 4.0 cm is placed vertically
in a uniform horizontal magnetic field of magnitude 1.0 T. If the current
flows in the coil is 5.0 A, determine the torque acting on the coil when
the plane of the coil is
a. perpendicular to the field,
b. parallel to the field,
c. at 60° to the field.
76
SF017
Solution: N=20
turns, A=24x10-4 m2,B=1.0 T, I=5.0 A
r
B o From the figure, θ =90° and φ =0° ,
θ = 90r
thus the torque on the coil is
A
τ = NIAB cos θ or τ = NIAB sin φ
τ =0
r
r
A
B From the figure, θ =0° and φ =90° ,
φ = 90 o thus the torque on the coil is
τ = NIAB cos θ or τ = NIAB sin φ
τ = 0.24 N m
a.
b.
r
B
c.
θ = 60 o
φ = 30
r
A
o
From the figure, θ =60° and φ =30° ,
thus the torque on the coil is
τ = NIAB cos θ or τ = NIAB sin φ
τ = 0.12 N m
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6.14 Hall Effect
Definition – is defined as the production of a potential difference
within a conductor or semiconductor through which a
current flowing when there is a strong transverse
magnetic field.
6.14.1 Explanation of Hall Effect
{
{
Consider a flat conductor (such as copper) carrying a current I in the
direction of +x-axis and is placed in a uniform magnetic field B (-z axis)
perpendicular to the plane of the conductor as shown in figure 6.14a.
+y
VH
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r
B
h
P
− − − − r− − r− − − −− − d
vd FB
I
I +x
r
−r
E
F
+ + + + + + + E+ + ++ +
Q
r
d : width of the conductor
B
h : thickness of the conductor
+z
Fig. 6.14a
78
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{
In metal the charge carrier is electron. The electrons drift with a drift
velocity vd in the opposite direction of the current I (shown in figure
6.14a).
{
{
{
The magnetic force FB acts on the electron in the direction upwards
(Fleming’s right hand rule) and cause the electron deflected to the upper
surface (P).
As time passes, more and more electrons will accumulate on the upper
surface (P) and left behind positive charges at the lower surface (Q).
This results in an electric field E acting in the direction upwards and the
electrons will experience electric force FE in the direction downwards.
{
{
The electric force FE will gradually increase as more electrons
accumulate at the upper surface.
An equilibrium will be reached when the magnitude of the electric force
FE becomes equal to the magnitude of the magnetic force FB and then
{
the electrons will drift along the –x axis without any more deflection.
At this instant, the upper surface (P) will be negative potential and the
lower surface (Q) will be positive potential. Hence a potential difference
will exist and known as Hall potential difference (voltage) VH.
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79
{
If the flat conductor is replaced with the flat semiconductor and the
charge carrier is positive charge, explain the phenomenon of hall effect
in this flat based on figure 6.14b. (exercise)
+y
r
B
h
P
VH
I
d
+
+z
r
B
Q
d : width of the conductor
Fig. 6.14b
6.14.2 Equation of Hall Voltage , VH
{
{
h : thickness of the conductor
According to the figures 6.14a and 6.14b, the Hall potential difference
(voltage) is
VH = Ed
When the equilibrium is reached between the electric and magnetic
force then
FB = FE where FE = qE and FB = Bqvd sin 90 o
Bqvd = qE
SF027
I +x
and
E=
VH
d
80
SF017
V 
Bqvd = q H 
 d 
VH = Bvd d
{
(6.14a)
From the definition of the drift velocity vd,
I
where e = q
nAe
I
vd =
n(dh )q
vd =
{
and
Therefore eq. 6.14a can be written as
 I 
d
VH = B
 ndhq 
 1  BI
VH =  
 nq  h
where
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A = dh
(6.14b)
1
: Hall coefficient
nq
h : thickness of the flat conductor/semiconductor
n : number of charge carriers per unit volume 81
(charge carrier density)
6.14.3 Uses of the Hall Effect
{
The Hall effect can be used to
z determine the sign of the charge carriers in a conductor /
semiconductor.
z determine the density of the charge carriers in a conductor/
semiconductor.
z measure the magnetic flux density B that is known as Hall probe.
{
Example 23:
A current of 3.00 A flows through a piece of metal of width 0.800 cm and
thickness 625 µm. The metal is placed in a magnetic field of flux density
5.00 T perpendicular to the plane of the metal. If the Hall voltage across
the width of the metal is 24.0 µV, determine
a. the drift velocity of the electron in the metal.
b. the density of the conduction electron in the metal.
(Given charge of the electron, e= 1.60 x 10-19 C)
Solution: I=3.00
A, d=0.800x10-2 m, h=625x10-6 m,
B=5.00 T, VH=24.0x10-6 V.
a. When the equilibrium is reached between the electric and magnetic
force then
FB = FE
SF027
Bevd = eE
and
E=
VH
d
82
SF017
VH
Bd
vd = 6.00 x10 −4 m s −1
vd =
b. By applying the equation of the Hall voltage, thus
 1  BI
VH =  
 ne  h
BI
n=
VH eh
n = 6.30 x10 28 electron m −3
{
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83
{
{
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Example 24: (exercise : unit 6.12)
A rectangular coil of 10 cm x 4.0 cm in a galvanometer has 50 turns and
a magnetic flux density of 5.0 x 10-2 T. The resistance of the coil is 40 Ω
and a potential difference of 12 V is applied across the galvanometer,
calculate the maximum torque on the coil.
Ans. : 3.0 x 10-3 N m.
Example 25: (exercise)
An electron moving at a steady speed of 0.50 x 106 m s-1 passes
between two flat, parallel metal plates 2.0 cm apart with a p.d. of 100 V
between them. The electron is kept travelling in a straight line
perpendicular to the electric field between the plates by applying a
magnetic field perpendicular to the electron’s path and to the electric
field. Calculate :
a. the intensity of the electric field.
b. the magnetic flux density needed.
Ans. : 0.50 x 104 V m-1, 0.010 T
Example 26: (exercise)
A moving coil meter has a 50 turns coil measuring 1.0 cm by 2.0 cm. It is
held in a radial magnetic field of flux density 0.15 T and its suspension
has a torsional constant of 3.0 x 10-6 N m rad-1. Find the current is
required to give a deflection of 0.5 rad.
Ans. :1.0 mA
84