Download magnetic effect of electric current

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Prof. Deepak Jawale
MAGNETIC EFFECT OF ELECTRIC CURRENT
Introduction
In this chapter we shall consider the various instruments for measurement of electrical
quantities such as current, potential difference, resistance etc. and the working of all these instruments is
based upon the magnetic effect of electric current.
A galvanometer is an instrument that detects electric current as well as indicates the
direction of current in the circuit. There are two main types of galvanometer.
i)
Moving coil galvanometer and
ii)
Moving magnet galvanometer or tangent galvanometer.
In a m.c.g. the coil is suspended in a strong magnetic field moves when the electric current
is passed through it. While in a Tangent galvanometer the coil is fixed and the magnetic needle moves.
Q.1
Describe the theory, construction and working of a moving coil galvanometer.
Construction (Suspended type MCG)
a)
b)
Principle : When a current carrying coil is suspended in a uniform magnetic field, a torque acts on it, due
to which the coil rotates. This rotation is opposed by a restoring torque produced by the suspension
system. When the two torques are equal, the coil is deflected by the angle . The deflection of the coil of
the moving coil galvanometer is directly proportional to the current flowing through the coil.
1)
Horse shoe magnet : A strong horse shoe magnet with concave pole pieces is used to produce a
radial magnetic field.
2)
Coil : A rectangular coil of thin insulated copper wire is suspended between the poles of the horse
shoe magnet. It is free to rotate about a vertical axis. The coil is suspended by means of a thin
phosphor-bronze fibre. The lower end of the coil is connected to a spring. The current enters the
coil through the fibre and leaves through the spring.
3)
Iron core : A soft iron cylinder is fixed, in the gap of the coil. The coil can freely rotate around
the cylinder. As the permittivity of soft iron is high, the iron core increases the strength of the
radial magnetic field.
4)
Lamp-scale arrangement : A small mirror is fixed to the suspension fibre. A lamp and scale is
arranged in front of the mirror. The reflected bright spot gives the deflection.
Theory and working :
1)
Consider a rectangular coil PQRS of length l and breadth b. The coil is suspended in a uniform
magnetic field of induction B. Let I be the current passing through it.
2)
Suppose the plane of the coil is parallel to the direction of the magnetic field. As the sides PS and
QR are parallel to the field, they do not experience force. The sides PQ and SR are perpendicular
to the direction of the field. The force acting on them is.
F = Bil
According to Fleming’s left hand rule, it acts normally outwards on PQ while normally inwards on
SR. (Due to the opposite directions of the currents)
3)
The two equal and opposite forces separated by the distance b form a couple given by
Torque = force x Perpendicular distance between the two forces
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5)
Prof. Deepak Jawale
 = Bil x b
 = BiA
(As l x b = A = area of the coil)
If the coil is having n turns, then
 = n BiA
This torque will rotate the coil.
A radial magnetic field is used in MCG. Therefore, the field remains parallel to the plane of the
coil in any position. Thus the torque,  = n BiA is constant in any position of the coil.
Due to the rotation of the coil the fiber gets twisted. A restoring torque acts on the coil. It is
proportional to the angle of twist.
Suppose the coil comes to rest at an angle of twist . Then, in this position,
Deflecting torque = restoring torque
nBiA = k 
where k is restoring torque per unit twist.
K
 i = ––––– 
n BA
As, k, n, B and A are constant.
I
Thus, the deflection of the coil in a moving coil galvanometer is directly proportional to
the current flowing through the coil.
Remark : Construction pivoted type MCG
1)
2)
3)
4)
A coil is placed between the two poles of horse shoe
magnet. The coil is mounted on a pivot, between two
supports. The supports are having jeweled bearings,
which are almost frictionless.
Two hair springs, one above and one below the coil,
control the rotation of the coil. The two springs are
spiraled in opposite direction. The current enters
through one spring and leaves through the other.
A long thin pointer is attached to the coil. It shows
the rotation of the coil on a scale.
The whole apparatus is fitted in a box having a
window through which the deflection can be
observed.
Comparison
1)
This pivoted type MCG is less sensitive as compared to the suspended type.
2)
It is portable and easier to use.
Q.2
Why should the magnetic field in a moving coil galvanometer be radial?
Ans : The radial field ensures that in any position the plane of the coil is parallel to the field and the
current flowing through the coil is directly proportional to the deflection of the coil. i.e., we can have a
linear scale. If this is not done, the angle made by the plane of the coil with the magnetic induction will go
on changing. This results in a varying torque acting on the coil, as the coil rotates. Therefore, there shall
not be a simple relation between the current and the deflection.
Q.3
Explain the term sensitivity of a galvanometer and explain how the sensitivity of a MCG can
be increased?
Ans : Sensitivity of MCG : It is defined as the ratio of the change in the deflection of the galvanometer
to the change in the current.
d
Sensitivity = –––
di
A galvanometer is said to be sensitive, if it gives a large deflection for a small current.
d

d
––– = –– if ––– is constant
di
i
di
For MCG, we have
k
i = ––––– 
nBA
nBA
 = ––––– i
k
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Differentiating the equation w.r.t. i
d nBA
––– = ––––
di
k
Hence, the sensitivity can be increased by,
1)
Increasing the number of turns (n) of the coil,
2)
Increasing the induction (B) of the magnetic field,
3)
Increasing area (A) of the coil,
4)
Decreasing torque per unit twist (k) of the fibre.
Limitations
1)
If a powerful magnet is used to increase B, it increases the weight of the instrument.
2)
Increase in n and A increase the weight of the coil. Hence, it requires a thick suspension fiber,
which will have a large value of k. Therefore, the sensitivity can only be increased up to a certain
limit.
Q.4
Ans :
Discuss the accuracy of MCG
The relative error in the measurement of the current i is given by the ratio di/i.
We have
K
i = ––––– 
n BA
Differentiating the equation w.r.t. 
K
di
d
 di = ––––– d ––– = –––
n BA
i

Hence, error in the measurement of the current depends only on the error d in the
measurement of the deflection.
For the greater accuracy in the measurement of the current, the ratio di/i should be small.
From the above relation, it is clear that the ratio will be small if the deflection is large for a given current.
Q.5
Explain why a) the resistance of an ammeter should be low. b) the resistance of a voltmeter
should be high.
Ans :
a) The ammeter is to be connected in series with other components (LCR) to read the current. If
the resistance of the ammeter is appreciable, it will change the current in the circuit. The
ammeter will then show a less value for the current.
b) The voltmeter is to be connected in parallel across the circuit element (LCR). If the resistance
of the voltmeter is not very high, it will draw current from the circuit element and the current
in the circuit element will be reduced. As a result, the voltage across the circuit element will
drop and voltmeter will show correspondingly less voltage. Ideally, the resistance of the
voltmeter should be infinity.
Q.6
Explain how will you convert a moving coil galvanometer into an ammeter. Derive the
necessary formula.
Ans :
1)
Following are the limitations to use a galvanometer directly to measure the current.
To measure a current flowing in a circuit, the galvanometer should be connected in series with the
circuit. In that case the resistance of the galvanometer is added to the circuit resistance. The
resultant current decreases and an error is generated in the measurement.
A galvanometer gives a full scale deflection for a very small current. Hence, a large current can
not be measured by it.
Also, the coil of a galvanometer is delicate and may be damaged with a large current.
2)
3)
To remove all these limitations
1)
2)
3)
The galvanometer resistance should be reduced.
There should be some arrangement to keep the deflection in the scale for a large current also.
It should be protected from the damage due to the large current.
This can be achieved by connecting a low resistance in parallel with the galvanometer
called shunt resistance.
1)
Let G be the resistance of the galvanometer coil.
2)
ig be the current flowing through the coil for full scale deflection.
3)
S be a shunt, a low resistance connected in parallel with the galvanometer.
4)
To measure a large current I, the shunt S is adjusted such that only the part i g passes through G and
the remaining part is flows through S. Then,
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
i = ig + i s
is = i – ig
Prof. Deepak Jawale
--------(1)
By Ohm’s law


igG = isS
igG = (i – ig)S
--------(2)
ig
––––– G
S=
--------(3)
i – ig
Hence, the required shunt to measure a current up to the maximum value i can be
calculated by the equation (3), and the scale is calibrated in terms of amperes.
Q.7
Explain how will you convert a moving coil galvanometer into a voltmeter. Derive the
necessary formula.
Ans :
Following are the limitations to use a galvanometer directly to measure the voltage.
1)
To measure a potential difference, it is necessary to connect the galvanometer in parallel with the
circuit. In that case some current flows through the galvanometer and the circuit current decreases;
hence the P.D. across the circuit element decreases and an error is generated in the measurement.
The galvanometer gives a full scale deflection when a very small P.D. is applied to it, hence the
large voltage can not be measured by it.
The large voltage applied across the galvanometer may pass a large current through its coil and the
coil may be damaged.
2)
3)
To remove all these limitations :
1)
2)
3)
The galvanometer resistance should be increased so
that a very small current will flow through it.
There should be some arrangement to keep the
deflection in the scale for a large voltage.
It should be protected from the damage.
This can be achieved by connecting a very high resistance in series with the galvanometer coil.
1)
2)
3)
4)
Let G be the resistance of the galvanometer.
ig be the current flowing through the coil for a full scale deflection.
Let R be a high resistance connected in series with the galvanometer.
To measure a large P.D. up to the maximum value V, the resistance R is adjusted such that the
current ig passes through the coil of the galvanometer at the potential V.
By Ohm’s law.
V = igG + igR
V
 R = ––– – G
ig
Hence, the required series resistant R to measure a potential upto the maximum value V
can be calculated by the above equation and the scale is calibrated in terms of volts.
Ampere’s law and its application

Ampere’s law : The line integral of the magnetic field ( B ) around any closed path is
equal to 0 times the total current (I) passing through that closed path.



 B . dl =  I
0
….. (1)
This is the mathematical statement of Ampere’s law. The quantity 0 represents the
absolute permeability of free space i.e. vacuum.
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Application of Ampere’s law :
1) Magnetic field due to a straight conductor carrying a current :
Consider a long, straight conductor carrying a current I. Let us determine the magnetic

induction B at a perpendicular distance r from the conductor, as shown in Fig. (a). For this purpose, let us
consider a circular Amperian loop of radius r, drawn in a plane perpendicular to the straight conductor, as
shown in Fig. (b). Note that in this figure, the direction of the current is outwards at right angles to plane
of the circle.
Fig. : a) Straight conductor carrying current I
(b) A circular Amperian loop in a plane perpendicular to the straight conductor

At every point of the loop, the magnetic induction B is directed in the tangential


direction. therefore the angle between B and the length element dl is zero at all the points.
The line integral along the closed loop is given by
 
 
 
B . dl =
B dl cos 0o = B dl = B dl




The total length of all the elements like dl is equal to the circumference of the circular loop.




dl = 2  r

 
B . dl = B (2  r)
….. (2)
According to Ampere’s law,
 
B . dl =  0 I

 B (2  r) = 0 I

B=
0 I
2r
….. (3)
This expression gives the magnitude of the magnetic induction at a distance r from a long
straight conductor carrying a current I.
Magnetic field due to long solenoid
Consider a long coil of wire consisting
of closely packed loops called as solenoid. Figure (a)
shows the magnetic field due to a solenoid carrying
current. It should be noted that magnetic field resembles
that of a bar magnet. Inside the solenoid, the magnetic
field is uniform and parallel to the solenoid axis. Outside
the solenoid, the magnetic field is very small as
compared to the field inside and may be assumed zero.
Fig. : Magnetic field due to long solenoid
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Consider a long solenoid having closely packed coils. Let I be the current through the
solenoid. Let n be the number of tuns per unit length of solenoid. Let B be the magnitude of magnetic
field inside the solenoid.
In order to use Ampere’s circuital law to determine the magnetic field inside a solenoid,
consider a rectangular closed path PQRS as shown in figure (b) where PQ = l.

The line integral of B over the closed path PQRS is given by,
 
Q

 
R

 
S

 
P

 
 B . dl = B . d l + B . d l + B . d l + B . d l
P
Q
R
S
Now,
  Q
B
 . dl =  B . d l cos 0 = B l
Q
P
P
Also,
P 

 
B
.
d
l
=
B
.
d

 l =0

R
Q

( angle between B and d l is 90o)
S
and
S
 
B . d l = 0
( field outside the solenoid is zero)
R

 
 B . dl
….. (i)
=Bl
According to Ampere’s circuital law, we have,
 
 B . dl = 0 x current enclosed by PQRSP
= 0 x number of turns in PQRSP x I

 
 B . dl
= 0 nl I
….. (ii)
From equation (i) and equation (ii), we have,
B l = 0 nl I

B = 0 nI
….. (iii)
Thus, equation (iii) gives magnetic induction due to a long solenoid.
Remark :
i)
It should be noted that, magnetic induction depends upon n and I. It does not depend upon the
position within the solenoid. Therefore, magnetic field inside the solenoid is uniform.
ii)
At points near the ends of the solenoid, the magnitude of magnetic field is
B=
1
2
0 nI
Magnetic field due to toroid
An endless solenoid in the form of a
ring is called a toroid. Figure shows the magnetic field
due to a toroid carrying current. The magnetic lines of
force inside the toroid are circular, concentric with the
centre of the toroid. It should be noted that magnetic
field inside the toroid is uniform because the toroid has
no ends.
Magnetic Effect of Electric Current
Fig. : Magnetic field due to toroid
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Consider a thin toroid having a large radius. Let I be the current through the toroid. Let a
be the mean radius of toroid. Let n be the number of turns per unit length. Let B be the magnetic field
inside the toroid.
In order to apply Ampere’s circuital law, we choose the closed path (inside the turns) as a
circle of radius a (dotted line path). Note that magnitude of B is the same everywhere on this closed path.



Further, angle between B and dl is zero everywhere on this path. Therefore, line integral of B over this
closed path is given by,
 
 B . dl =  B.dl cos 0 = B 2  a
 
 B . dl = B 2  a
….. (i)
According to Ampere’s circuital law,
 
 B . dl = 0 x current enclosed by closed path
 
or
 B . dl = 0 x n (2  a) I
From equation (i) and (ii), we have,
….. (ii)
 Current enclosed = n x length of path x current 
 by closed path = n x 2  a x I



B 2  a = 0 x n x 2  a x I
B = 0 nI
….. (iii)
Thus, equation (iii) gives magnetic induction due to a toroid.
Remark :
1)
It should be noted that magnetic field inside the toroid is independent of the radius of toroid. It is
because magnetic field is only confined to inside the toroid.
2)
The magnetic field inside a toroid is constant and is always tangent to the field lines.
Cyclotron :- Cyclotron is a device used to accelerate positively charged particles to acquire enough
energy to carry out nuclear disintegrations.
Principle :- When a positively charged particle is made to move in periodic time again and again in a
high frequency electric field and using strong magnetic field, it gets accelerated and acquires sufficiently
large amount of energy.
Construction :- It consists of two hollow D-shaped
metallic chambers D1 and D2 called dees. These dees are
separated by a small gap where a source of positively
charged particles is placed. Dees are connected to high
frequency oscillator, which provides high frequency
electric field across the gap of less. This arrangement is
placed between two poles of a strong electromagnet. The
magnetic field due to this electromagnet is perpendicular
to the plane of dees as shown in figure.
Fig. : Cyclotron
Working : If a positively charged particle is emitted from O, when dee D2 is negatively charged and dee
D1 is positively charged, it will accelerate towards D2, it is shielded from the electric field by the metallic
chamber. Inside D2, it moves at right angle to the magnetic field and hence describes a semicircle inside
it. After completing the semicircle, it enters the gap between the dees at the time when polarities of dees
have been reserved. Now the charge particle is further accelerated towards D1. Then it enters D1 and again
describes the semicircle due to the magnetic field which is perpendicular to the motion of the proton. This
process continues till the proton reaches the periphery of the dee system. At this stage, the particle is
deflected by the deflecting plate which then comes out through the window (W) and hits the target.
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Theory :
Prof. Deepak Jawale

When a particle moves at right angles to the magnetic field ( B ) inside the dee, magnetic
force acting on it is,
F = B q v sin 
F=Bqv
This force provides the centripetal force mv2/r to the charged particle to move in a
circular path of radius r.

Bqv =
or
r=
mv 2
r
mv
….. (i)
qB
Time taken by the particle to complete the semicircle inside the dee,
distance
t=

speed
r

v


v
x
mv
qB
m
t=
….. (ii)
qB
This shows that time taken by the positively charged particle to complete any semicircle is same.
1)
Time period : Let T be the time period of the alternating electric field, then polarities of dees will
change after time T/2. The particle will be accelerated if time taken by it to describe semicircle is
equal to T/2.
T
e.g.
=t=
2
qB
m
T=
2)
m
….. (iii)
qB
Cyclotron frequency : Cyclotron frequency which is also called as magnetic resonance
frequency is given by
 =

1
T
=
qB
2 m
Cyclotron angular frequency is given by,
=2=
3)
….. (iv)
qB
m
….. (v)
Energy gained : Energy gained by the positively charged particle is given by,
E=
Also, v =

1
2
mv2
qBr
m
E=
E=
1
2
m
q 2 B2 r 2
m2
q 2 B2 r 2
2m
….. (vi)
Thus, maximum energy gained by the charged particle is
Emax =
q 2 B2
2m
2
rmax
….. (vii)
Thus, the positively charged particle will acquire maximum energy when it is at the
periphery of the dees.
Limitations of cyclotron :
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1)
Cyclotron can not accelerate uncharged particles like neutron.
2)
Cyclotron can not accelerate electrons they have very small mass. Electrons start moving at a
very high speed when they gain small energy in the cyclotron. Oscillating electric field makes
them to go quickly out of step because of their very high speed.
Important Formulae
1)
Torque acting on a coil placed in a magnetic field.  = nBiA cos  (Where , is the angle between
the plane of the coil and the direction of B).
2)
For a MCG, nBiA = k
 K 

 n BA 
 i= 
3)
For a tangents galvanometer B = BH tan 
0 n
2r
= BH tan 
 2rBH 
 tan 
 0 n 
 i = k tan 
 i= 
Where, k =
2rBH
0 n
is constant and is called the reduction factor of TG
4)
Sensitivity of MCG, S = S =
5)
Sensitivity of TG,
S=
6)
nAB
k
0 n cos 2 
2rBH
The suitable shunt resistance in an ammeter
 ig

 G
 i  ig 
S= 

7)
The suitable series resistance in a voltmeter.
 R=
V
ig
G
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