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EaI svaamaI ivavaokanaMd iSaxaNa saMsqaa kaolhapUr saMcalaIt
ivavaokanaMd ka^laoja¸kaolhapUr
XI SCIENCE
[PHYSICS – II]
MAGNETIC EFFECT OF
ELECTRIC CURRENT
Prof. Shri. R. S. Gade
January 30, 2013
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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MAGNETIC EFFECT OF ELECTRIC CURRENT (9)
[SYLLABUS]
1.
2.
3.
4.
5.
6.
7.
8.
Biot - Savart law: Statement and explanation of magnetic field due to current element.
(Derivation of magnetic induction due to a straight conductor carrying a current is not
expected.)
Right hand rule: Statement (Thumb rule)
Magnetic induction at the centre of circular coil carrying current: Derivation, concept of
polarity of face of the coil.
Magnetic induction at a point along the axis of a coil carrying current: Derivation.
Fleming’s left hand rule: Statement.
Force between two infinitely long current carrying parallel conductors: Derivation. (In case of
same and opposite directional currents)
Definition of Ampere: Definition.
Force acting on a conductor carrying current in magnetic field: Derivation.
* * * * *
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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MAGNETIC EFFECT OF AN ELECTRIC CURRENT
Introduction: When an electric current flows through a conductor the magnetic field is produced
around that conductor this is known as magnetic effect of electric current. The strength of this
magnetic field is measured in the form of magnetic induction which can be defined as: “The
magnetic induction of a magnetic field at any point can be defined as the magnetic flux per
unit area around that point”. The SI unit of magnetic flux is weber, hence SI unit of magnetic
induction is Wb/m2 known as tesla.
Biot - Savart law (Laplace’s law):
Statement: “The magnetic induction at a point due to an element of a conductor carrying a
current is: (i) directly proportional to the current, (ii) Directly proportional to length of
element, (iii) Directly proportional to sine of the angle between the element and the line
joining the centre of element to the point and (iv) inversely proportional to the square of the
distance of the point from the centre of element”
Explanation: Consider a small element dℓ of a conductor carrying current I. Let a point P is at
a distance r from the centre of the element on a line making an angle of θ with element. Then
according to Biot - Savart law the magnetic induction dB at the point P due
I dℓ sin θ
I dℓ sin θ
dB
to current carrying element dℓ is: dB ∝
∴ dB = K
---- (i)
2
r
r2
I
where K is the constant of proportionality depends upon the system of units
used and the medium surrounding the conductor. In the SI system this
θ r
μ
dℓ
constant for vacuum of air is 0 where μ0 is permeability of vacuum.
4π
μ I dℓ sin θ
Therefore equation (i) can be written as: dB = 0
------- (ii)
4π
r2
Equation (ii) gives the magnitude of magnetic induction at a point P. Its
direction can be find out by using right hand rule.
*
By using this law we can prove that magnetic induction due to a current carrying straight
μ I
conductor at appoint distance r from it is given by: B = 0 .
2π r
*
Right hand rule: Hold the current carrying conductor in right hand with fingers curled around
it and thumb stretched along its length in the direction of current then curled fingers indicates
the direction of the magnetic field.
Magnetic induction at the centre of a circular coil carrying current:
Consider I is the current flowing through a circular loop of radius r in anticlockwise
direction as shown in fig. To find magnetic induction at the centre of loop, let loop is divided
into small elements each of length dℓ. Consider one such element at point
dℓ
•
A. According to Biot-Savart law, the magnetic induction dB at the centre
A
I
μ I dℓ sin θ
r
O of the loop, due to this element is given by: dB = 0
4π
I
I
O
But for any element θ = 900 so that sinθ = 1
μ I dℓ
∴ dB = dB = 0 2 ----- (i)
4π r
∴ The total magnetic induction due to entire loop is obtained as: B = dB
∴B=
∴B=
μ0 I dℓ
4π
μ0 I
4π r 2
r2
=
2πr =
μ0 I
4π r 2
μ0 I
2 r
dℓ but
r2
dℓ = circumference of loop = 2πr
----- (ii)
μ
nI
If the circular loop has n turns each of radius r then: B = 0
----- (iii)
2 r
This is the magnitude of induction at the centre of the circular coil. It is directed at right angles
to the plane of coil as decided by right hand rule.
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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*
The current carrying coil is equivalent to a bar magnet (i.e. magnetic dipole), with one face
acting as N-pole and other as S-pole according to following diagrams. Which proves that, The
face through which
current appears to
flow in anticlockwise
direction acts as N•
pole while the face
through
which
current appears to
FaceB
Face A
Face A
Face B
flow in clockwise
A
direction acts as S-pole of the equivalent bar magnet. The magnetic dipole moment for such
coil is (nIA).
Magnetic induction at a point along the axis of a circular coil carrying a current:
Consider a circular loop of wire of radius a carrying a current I. Let P be a point on the
axis of the loop at a distance of x from the centre of the loop. To find the magnetic induction at
point P suppose loop is divided into even number of elements
A
each of length dℓ. Consider one such element at the point A.
S
R
Join AP. Let AP = r and ∠ APO = α. Then magnetic induction
r
a
dB at point P due to element at A is given by Biot-Savart law
α
μ I dℓ sin θ
I
α
Q
as: dB = 0
. But AP is perpendicular to element so
O
2
’
x
4π
Q
P
r
μ I dℓ
0
that θ = 90 and sinθ = 1. ∴ dB = 0 2 ---- (i)
4π r
’
’
This
induction
is
perpendicular
to
the element dℓ at A, and
S
R
the line AP and is represented by PR as shown in fig. The
B
vector PR can be resolved into two components PS
perpendicular to axis and PQ parallel to axis. From figure: PS = PR cosα and PQ = PR sinα.
In the same manner magnitude of dB is produced at P by the current in element dℓ at
point B, diametrically opposite to A. However its direction is perpendicular to element at B
and the line BP. Let this induction is represented by PR′ which can be resolved into PS ′ and
PQ′ as shown in figure.
If we consider the induction at point P due to all the elements of loop, it is found that the
components of induction perpendicular to axis cancel each other, while the components
parallel to axis are added. Thus resultant induction B at point P due to circular loop is the sum
of all the components parallel to the axis of loop. From figure: PQ = PR sinα = dB sinα.
μ0 I dℓ
∴ B = dB sinα =
sin α.
4π r 2
From ∆ AOP, sinα = a/r
μ0 I dℓ a
μ Ia
∴B=
= 0 3 dℓ
2
4π
r
r
4π r
I a2
μ
but dℓ = 2πa ∴ B = 0 3
2 r
But from ∆ AOP, r2 = a2 + x2 i.e. r3 = (a2 + x2)3/2
∴B=
μ0 I a 2
3
2 a 2 +x 2 2
.
If we have a circular coil of n turns then: B =
μ0 nI a 2
3
2 a 2 +x 2 2
This is required equation.
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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Force acting on a conductor carrying current in magnetic field:
If a charged particle carrying charge q is moving with velocity v in uniform magnetic
field of induction B it is acted by a force F given by: F = q v x B , which is perpendicular to
both direction of motion of charge and direction of magnetic field.
Now if we consider a
conductor carrying current I
𝐅
𝐅
and is placed in a magnetic
𝐁
𝐁
induction B then current in
conductor is due to electrons
θ
θ
moving in a direction
𝐯
I
opposite to direction of
current. Let (-dq) be the
Force acting on charge moving
Force acting on current
charge on electrons passing
in uniform induction
carrying conductor
through an element dℓ of the
conductor in time dt. We can assume that in time dt a positive charge (+dq) moves through a
distance dℓ through conductor in the direction of current so that, velocity of charge v =
∴ Force acting on the charge dq is: dF = dq v x B = dq
dℓ
dt
xB =
dq
dt
dℓ
dt
.
dℓ x B
dq
∴ dF = I dℓ x B
∵ =I
dt
As the charge is moving through conductor the force acting on the length dℓ of the conductor
is given by above equation i.e. dF = I dℓ x B
∴ Total force acting on the conductor of length ℓ will be:
F = dF = I dℓ x B = I
dℓ x B = I ℓ x B
∴ The magnitude of the, F = I ℓ B sinθ where θ is the angle between B and ℓ .
The direction of this force is given by Fleming’s left hand rule.
*
Case – I When the conductor is kept perpendicular to B then θ = 900 or 1800 therefore sin θ = 0
hence F = 0. Thus a straight conductor carrying a current does not experience a force when it is
kept parallel to B.
Case – II When the conductor is kept perpendicular to B then θ = 900 or 2700 therefore sinθ = 1
hence F = I ℓ B. Thus a straight conductor carrying a current experiences a maximum force
when it is kept perpendicular to B.
*
As B = F / I ℓ, SI unit of B is N/Am called tesla or Wb/m2. CGS unit is dyne emu-1 cm-1 called
oersted or gauss.
*
Lorentz force: Static charge produce an electric field, the current or moving changes produce
an additional a magnetic field which is a vector field. hence if we consider the point charge q
moving with velocity V and located at a distance r at a given time t in presence of both
magnetic and electric field is given by:
F = q E + q V x B = Felectric + Fmagnetic , at a distance r. This force is called as
Lorentz force.
Fleming’s left hand rule: “Stretch the fore-finger, middle finger and the thumb of left hand
mutually perpendicular to each other. Now hold the hand such that fore-finger denotes
direction of field, middle finger denotes the direction of current, and then thumb denotes the
direction of force acting on the conductor”
Force between two infinitely long current carrying parallel conductors:
(a) Like currents: Consider two very long straight conductors kept parallel to each other in a
horizontal plane, carrying currents I1 and I2 respectively in the same directions as shown in
figure. Let ℓ be length of each conductor and a is the distance between them. The first
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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conductor produces a magnetic field of induction B1 at every points of the second conductor.
By the right hand rule B1 is directed vertically downwards and its magnitude
μ I
a
is: B1 = 0 1 .
𝐁𝟐
2πa
The second conductor carrying a current I2 is perpendicular to this induction
B1 and hence acted upon by a force F2 given by:
𝐅𝟏 𝐅𝟐
𝐁𝟏
μ I
μ I I ℓ
F2 = I2 ℓ B1 = I2 ℓ 0 1 = 0 1 2 ------ (i)
2πa
2πa
According to Fleming’s left hand rule this force is horizontally towards
first conductor.
𝐋𝐢𝐤𝐞 𝐜𝐮𝐫𝐫𝐞𝐧𝐭𝐬
In the same manner the second conductor produces magnetic induction B2
at every point of first conductor, which according to right hand rule directed vertically upwards
μ I
and its magnitude is given by: B2 = 0 2 .
2πa
The first conductor carrying current I1 is perpendicular to this induction B2 and thus acted
I1
I2
μ I
μ I I ℓ
upon by a force F1 given by: F1 = I1 ℓ B2 = I1 ℓ 0 2 = 0 1 2 ------- (ii)
2πa
2πa
According to Fleming’s left hand rule this force is horizontally directed towards the
second conductor.
μ I I ℓ
Thus, the two conductors attract each other with the same force given by: F = 0 1 2
*
μ 0 I1 I2
2πa
The force of attraction per unit length of each conductor is given by: f =
.
2πa
(b) Unlike currents: As we have studied the case of like currents it can be proved that parallel
conductors carrying currents in opposite directions repel each other
a
with a force of magnitude:
𝐁𝟐
μ I I ℓ
𝐁𝟏
F= 012
𝐅𝟏
𝐅
2πa
𝟐
I1
I2
𝐔𝐧𝐥𝐢𝐤𝐞 𝐜𝐮𝐫𝐫𝐞𝐧𝐭𝐬
Definition of ampere: The unit of current in SI system is called ampere and it can be defined
as follows:
We know when two long parallel straight conductors carrying either like or unlike
currents repels each other or attracts each other with a force per unit length is given by:
μ I I
f= 012
2πa
μ
In this if I1 = I2 = 1 A ; a = 1 m then, f = 0 = 2 x 10-7 N/m
2π
Thus, “A current of 1 ampere is that current which when flowing through each of two
very long straight parallel conductors kept 1 metre apart in free space, produces a force of 2 x
10-7 newton per metre length of each conductors”
Torque on a current carrying loop in magnetic field:
Consider a rectangular loop of wire ABCD of length ℓ, breadth b and carrying a current I
placed in a uniform magnetic field of induction B made by a concave shaped pole pieces of a
permanent magnet such that plane of coil is parallel to B as shown in fig (a). The sides BC and
DA are parallel to B so that no force acts on them. The forces F1 and F2 on the sides AB and
CD respectively constitute the couple which rotates the loop in the sense shown and has
magnitude: τ = (I ℓ B) x b = I B (ℓ x b) = I B A where A = ℓ b is the area of the loop. For coil
of N turns, τ = N I A B = M B. where M = N I A is the magnitude of the magnetic dipole
moment M of the current carrying coil.
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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As the loop rotates the angle between M and B reduces from 900. Fig (b) shows a position
of the coil when the angle between M and B is θ. In the figure ⊗ represents the current in the
arm AB is directed into the page and ʘ represents the current in the arm DC directed out of the
page. In this position the sides BC and DC are acted upon by the forces of equal magnitude,
opposite in direction and having the same line of action, hence resultant of these two forces is
zero. But forces acting on the sides BC and DA are again constitutes the couple with the
F2
𝐅𝟐
Z
Side CD °
(current out)
C
B
b sinθ
M
B
θ
D
I
θ
B
A
x
𝐅𝟏
Side AB
(current in)
Y
X
*
*
Fig (a)
Fig (b)
F1
moment arm b sinθ as shown in fig (b). Hence torque now becomes:
τ = N (Iℓb) x bsinθ = N IABsinθ = M B sinθ = M X B this is the
required expression for the torque acting on current carrying loop in a magnetic field.
Now when the plane of coil is perpendicular to B then torque is minimum i.e. zero but the flux
through the coil is maximum and when the coil is parallel to B then the torque is maximum i.e.
i.e. NIAB but the flux is minimum.
Thus a current carrying coil suspended in a magnetic field experiences a torque which rotates
the coil and tends to maximize the magnetic flux through the coil.
* * * * *
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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SOLVED PROBLEMS
1. A straight conductor wire carries a current of 5 A. Calculate the magnitude of magnetic
induction at a point 10 cm away from the conducting wire. (μ0 = 4π x 10-7 Wb /Am)
µ I
µ 2I
10 −7 x 2 x 5
Solution: B = 0 = 0 =
= 10-5 Wb/m2
2πa
4πa
0.1
2. A current of 10 A passes through a coil having 5 turns and produce a magnetic field at the
centre of the coil having magnitude 0.5 x 10-4 T. Calculate diameter of the coil. (μ0 = 4π x 10-7
Wb /Am).
µ nI
3.
4.
5.
6.
4π x 10 −7 x 5 x 10
Solution: B = 0 =
= 0.5 x 10-4 ⇒ r = 1.257 m
2r
2r
A straight current carrying conductor, 30 cm long carries a current of 5 A. It is placed in a
uniform magnetic field of induction of 0.2 T, with its length making an angle of 600 with the
direction of field. Find force acting on conductor. [sin 600 = 0.8660]
Solution: F = B l I sin θ = 0.2 x 0.3 x 5 x sin 600 = 0.2598 N
Two long parallel current carrying conductors are 0.4 m apart in air and carrying current 5 A
and 10 A respectively. Calculate the force per meter on each wire, if the currents are in the
same directions and in the opposite directions. (μ0 = 4π x 10-7 Wb /Am).
Solution: In both the cases the magnitude of the force per unit length is same and is:
µ 2I I
f = 0 1 2 = 10-7 x 2 x 5 x 10 / 0.4 = 2.5 x 10-5 N.
4πa
∴ When the currents are in the same direction then force of attraction per meter of wire is 2.5 x
10-5 N and when the currents are in the opposite direction then the force of repulsion per meter
of each wire is 2.5 x 10-5 N
A particle carrying a charge moves with velocity 3 x 106 m/s at right angles to uniform field of
magnetic induction 0.005 T. Find the value of the charge on the particle if it experiences a force
of 2 x 10-2 N.
Solution: F = q v B sinθ = q v B ⇒ q = F / v B = (2 x 10-2) / 3 x 106 x 5 x 10-3 = 1.333 x 10-6 C.
Calculate the magnetic field of induction, due to circular coil of 400 turns and of radius of 0.05
m carrying a current of 5 A, at a point on the axis of coil at a distance of 0.1 m. (μ0 = 4π x 10-7
Wb /Am).
Solution: B =
µ 0 n I a2
3
2 a2 + x2 2
=
4 x 3.14 x 400 x 0.05 2
3
2 0.05 2 + 0.1 2 2
=
4 x 22 x 5 x 10 −7
2x7
3
25+100 x 10 −4 2
= 2.248 x 10-3 T
7. A conductor of length 25 cm is placed (a) parallel (b) perpendicular (c) inclined at an angle of
300 to uniform field of magnetic induction 2 T. If a charge of 1C passes through it in 5 s,
calculate the force experienced by the conductor.
Solution: (a) Sinθ = 0 ∴ B = 0
(b) Sin900 = 1 ∴ F = q v B = 1 x 5 x 10-2 x 2 = 0.1 N
(c) sin 300 = 0.5 ∴ F = q v B x 0.5 = 1 x 5 x 10-2 x 2 x 0.5 = 0.05 N
8. A proton from cosmic rays enters the earth’s magnetic field in a direction perpendicular to the
field. If the velocity of proton is 2 x 107 m/s and B = 1.6 x 10-6 Wb/m2, find the force exerted on
the proton by the magnetic field. (Charge on electron, e = 1.6 x 10-19 C)
Solution: The magnetic force on the proton, F = e V B = 1.6 x 10-19 x 2 x 107 x 1.6 x 10-6
= 5.12 x 10-18 N
9. A circular loop has a radius 10 cm and it is carrying a current of 0.1 A. Calculate its magnetic
moment.
Solution: The magnetic moment, M = N I A = N I π R2 = 1 x 0.1 x 3.142 x 0.1 x 0.1
= 3.142 x 10-3 Am2
* * * * *
[XI PHY - II/MAG. EFF. ELEC. CURR./ RSG]
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