Download MAGNETISM - auroraclasses.org

Magnetic field of electric current
Magnetic fields
Objects released from rest fall towards the earth. If a mass located at a certain position experiences a
force and undergoes a linear acceleration as a result, we conclude that there is a gravitational field acting in
that direction at that location.
Raindrops fall because of the earth's gravitational field.
Similarly, if an electric charge experiences a force in a region, we conclude that there is an electric field
acting in the region. The electric charge would undergo linear acceleration in that region.
Electric field near a charged comb attracts pieces of paper
Just as a gravitational field is a region in space where a mass experiences a force, and an electric field
is a region in which an electric charge experiences a force, a magnetic field is a region in which a magnetic
pole experiences a force.
Region of magnetic field exists around a magnet
The pins become magnets by induction. They have two opposite poles of equal strength. All magnets
are therefore also called magnetic dipoles.
In the region around the horse-shoe magnet shown, its magnetic field is non-uniform. That is, the field
is strong at positions closer to the poles than at distant points. In such non uniform fields the induced equal
poles at the two ends of the pin experience unequal forces. Thus, in such non-uniform fields the pins
experience a net force and undergo linear acceleration.
The earth creates a magnetic field around itself. This field is uniformly strong in any one region. In such
uniform magnetic fields a magnetic dipole does not experience linear acceleration. It is a matter of common
experience that a magnetic needle suspended or pivoted does not try to move itself towards the north of the
room. It merely turns to settle in a north-south direction. Even if disturbed from this orientation, the magnetic
dipole will repeatedly come to rest pointing in the same direction.
Thus if a magnetic dipole, after oscillating for awhile, persistently settles pointing in the same direction,
we may conclude that there exists a magnetic field in the region. In fact it is because a magnetic needle fitted
into a compass always comes to rest pointing in the same direction, (which we call the magnetic north,) we
conclude that there is a magnetic field around the earth.
Oersted's experiment
Experiment shows that in the vicinity of a stationary electric charge, an electric dipole aligns in a
particular direction but a magnetic dipole does not. Thus, there exists only an electric field but not a magnetic
field in the vicinity of a stationary electric charge. But when an electric charge is moving, (or if there is an
electric current) a magnetic needle placed near it sets always in a particular direction proving the existence of a
magnetic field.
Oersted placed iron filings on a glass plate having a small hole drilled in it. He had a wire passing
through the hole. When the wire was connected to a battery and the glass plate lightly tapped all the iron filings
arranged themselves in concentric circles centred at the point of intersection of the wire and the glass plate. If
the current strength is increased then the circles crowd nearer the wire.
A magnetic needle pivoted near the current is found to always settle tangential to the concentric circle
passing through that point. This indicates that there exists a magnetic field at every point in the vicinity of a
current carrying conductor. The needle settles in the direction of the magnetic field.
As indicated in the diagrams above, the direction in which the needle deflects is perpendicular to the
current and it is also perpendicular to the displacement of the point from the current carrying conductor.
On the axis of the current however, there is no magnetic field.
Direction of Magnetic Field
(right hand thumb rule)
The direction of this magnetic field created by the current in the conductor is perpendicular to the plane
of the paper which contains the current segment and the displacement of the location from it, and is given by
the right hand thumb rule as shown in the diagram above. The right thumb is directed along the current. The
palm is along the displacement of the location from the current. The fingers bent at right angles from the palm
indicate the orientation of the magnetic field.
Strength of Magnetic Field
The intensity of a gravitational field at a point is given by the force experienced by a unit mass placed at
that point. The strength of an electric field is likewise assessed by the force experienced by a unit positive
charge placed at that point. The strength of a magnetic field however is inconvenient to measure in a similar
manner because magnetic poles occur in inseparable pairs. Thus the net force on a dipole is zero in a uniform
magnetic field despite the field being of considerable strength.
Hence, the more commonly used index of the magnetic field is the number of lines of force per unit
area held at right angles. In a strong magnetic field there are many lines of force per unit area. Hence the
magnetic FLUX DENSITY (B) also called MAGNETIC INDUCTION is a more convenient measure of the
magnetic field. The units of magnetic flux density B are weber/metre2 (Wb/m2) which is also called Tesla (T)
Magnetic lines of force




A magnetic field is visualised using magnetic lines of force which are imaginary lines such that the
tangent at any point gives the direction of the magnetic field at that point.
Near an isolated magnetic dipole, they are continuous closed curves, being from the north pole to the
south pole outside the magnetic dipole and south to north pole internally.
Magnetic lines of force never intersect.
They are in a state of longitudinal tension and lateral repulsion.
If iron filings are sprinkled around a magnet resting on a board, on tapping the board the iron filings are
found to arrange themselves in well defined curves around the magnet.
These curves describe the lines of force around the magnet.
Behaviour of magnetic needle in magnetic field
Since a magnetic needle has two opposite poles of equal strength, it experiences two forces. If the field
is uniform (everywhere the same in magnitude and direction), then these two forces are oppositely directed
giving a net force zero.
These forces form a couple. A couple produces angular acceleration of the needle turning it about its
middle. The behaviour of this magnetic dipole in a uniform magnetic field is identical to the behaviour of an
electric dipole in a uniform electric field.
The dipole oscillates about a mean position aligned with the magnetic field and ultimately comes to rest
in this position.
If the magnetic field in which the dipole is placed is non-uniform, then the midpoint of the dipole moves with
linear acceleration even as the dipole oscillates.
If a magnetic needle placed at a point oscillates and finally comes to rest always in a particular
direction, then we can conclude that there is a magnetic field at that point.
Magnitude of flux density - Biot Savart Law
The flux density due to a current should depend upon the strength I of the current. Like in other fields,
due to a infinitesimally section of length dl (analogous to a point mass or point charge), it should be inversely
proportional to the square of the displacement r. Furthermore, since the current creates no flux density on its
axis, the location of the point P with reference to the direction of the current is also significant.
Thus, if the current vector Idl is resolved along the displacement vector and perpendicular to it, the
point P would lie on the axis of the current Idl cos . Hence only the component Idl sin would give rise to flux
at the point P.
Hence, the magnetic flux density dB caused by a current I at a displacement r due to an infinitesimally
small segment dl of conductor carrying the current I, is found to be directly proportional toIdlsin and inversely
proportional to r2, where is the angle between the direction of the current vector (Idl) and the displacement
vector (r).
This is known as Biot Savart Law and gives the magnitude of the flux density created by only the current
carrying segment dl of the conductor at the point P.
This law gives the flux density created by only an infinitesimally small length of the wire and cannot
therefore be experimentally checked, but derivations based upon this law for currents of regular and finite path
lengths are experimentally verifiable.
Magnetic induction at the centre of a flat circular coil carrying current
The figure above represents a flat circular coil of centre O and radius r placed in air and carrying
current I.
Consider an infinitesimally small length dl of the current carrying wire. The direction of the current in the
wire is tangential. The displacement of the centre O of the coil from this segment dl of the wire is the radius r of
the coil. The angle between the radius r and the current carrying segment Idl is thus a right angle.
We have, from Biot Savart Law
For all the infinitesimally small current carrying segments, the distance from the centre is the same (r)
and the angle between the radius and the tangent is always a right angle. Hence, the current I, the distance r
and the sine of the angle between the current I and the displacement r, are all constants.
We therefore have the expression for the flux density caused by any infinitesimal segment of the
current carrying wire as:
This flux density due to each segment is in the same direction as shown below.
Thus the entire flux density B, is found by integrating the expression for dB for the whole coil.
where n is the number of turns in the coil.
The flux density at the centre of a current carrying flat circular coil is thus directly proportional to the
current I it carries, and also to the number of turns n of the coil. It is inversely proportional to the radius of the
coil.
Every current carrying coil is a magnet
By applying the right hand thumb rule we find that the direction of the flux density is out of the plane of
the coil if the current is anticlockwise and it is into the plane of the coil in the case of a clockwise current.
The flux is perpendicular to the plane of the coil. In the case of a bar magnet the flux through the
magnet itself is perpendicular to its cross sectional area. The flux emerges from the magnet at its north pole
and re-enters the magnet at its south pole end.
The current carrying coil is therefore like a rod magnet or a bar magnet in terms of its flux pattern. That
face carrying current counter clockwise (from which the flux emerges from the coil) is the north pole face of the
coil. The clockwise current face (into which the flux enters the coil) is its south pole. The thickness of the coil is
the length of the effective magnet. A current carrying coil is therefore a MAGNETIC DIPOLE. The axis of the
dipole is perpendicular to the plane of the coil.
Magnetic induction on the axis of a current carrying flat circular coil
The flux density dB due to an infinitesimally small segment of length dl carrying current I at a
perpendicular distance r from it must be perpendicular to the plane containing the vectors Idl and r (i.e. the
shaded plane shown in the diagram below. The cosine components of all such flux densities due to the whole
coil would nullify as they point radially outwards from the axis of the coil. The sine components would all add
up to the resultant flux density B.
As such we have,
where a2 is the area A of the flat circular coil.
where P (= IAn) is the magnetic dipole moment of the current carrying coil and a is negligibly small in
comparison with x.
Note the similarity between this, the expression for flux density at the end-on position of a magnetic
dipole, and the corresponding expression
giving the electric field strength on the end-on position of an electric dipole.
Magnetic induction due to long straight conductor carrying current:
Let dl be an infinitesimally small segment of the long straight conductor carrying current I. Then the flux
density dB at the point P is given by Biot Savart Law as given below. The displacement vector of the point P
from the segment dl being r makes an angle with the current direction. Thus, for a segment at the lower
extreme of the very long conductor, the angle would be zero. While for the upper extreme the angle would
be as high as 180o.
The flux density created by an infinitely long straight conductor is thus directly proportional to the
current it carries and inversely proportional to the perpendicular distance of the point from the conductor.
Magnetic induction on the axis of an infinite solenoid
A coil whose radius a is negligibly small in comparison with the axial length x may be considered of
infinite length.
Flux density on the axis of a flat circular coil has already been derived earlier in the chapter. This is
given by:
where n = the total number of turns in the solenoid.
Consider a flat circular coil of length dx. If the solenoid has N turns per unit length, then the number of
turns in our flat circular coil of length dx would be given by
n = Ndx
Thus the flux density dB due to a flat circular coil at a point distant x from its centre and located on its
axis is given by
This is the value of flux density only along the axis and is directed along the axis from that end where
the current is clockwise to that in which the current is anticlockwise. Its magnitude depends on a property of
the medium of the core - namely its permeability. It also depends on properties of the coil - the current it carries
and the number of turns per unit length.
The contribution of the coil to the flux density is also calledmagnetizing field intensity H (=NI) having
units ampere turns per metre or Am-1.
Therefore,
B= 0H
If the current in the coil is varied, the value of H varies and the flux density (or magnetic induction) B in
the core varies with it.
Numericals
UNLESS OTHERWISE STATED, g
e = 1.6 x 10-19 C AND o = 4 x 10-7 SI UNITS

=
10
ms-2 AT
THE
SURFACE
OF
THE
EARTH,
A 1long, straight horizontal wire carries a current of 10.0 A from west to east. Find the magnitude and
direction of the magnetic flux density at a point on the ground 2.0 m directly below the wire. [1.0 x 10-6T]
i] 2

ii]
A long straight wire of radius a carries a steady current. Sketch a diagram showing the lines of
magnetic flux density B near the wire and the relative directions of the current and B. Describe,
with the aid of a sketch graph, how B varies along a line from the surface of the wire at rightangles to the wire.
Two such identical wires R and S lie parallel in a horizontal plane, their axes being 0.10 m apart.
A current of 10 A flows in R in the opposite direction to a current of 30 A in S. Neglecting the effect
of the earth’s magnetic flux density calculate the magnitude and state the direction of the magnetic
flux density at a point P in the plane of the wires if P is :
a.
midway between R and S,
b.
0.05 m from R and 0.15 m from S.
[1.6 x 10-4 T; 0]


3
A long air-cored solenoid has two windings wound on top of each other. Each has N turns per
metre and resistance R. Deduce expressions for the flux density at the centre of the solenoid when the
windings are connected to a battery of EMF E and negligible internal resistance
i]
in series,
ii]
in parallel
(In each case the magnetic fields produced by the two windings reinforce.)
[B = oNE/R; B = 2 oNE/R]
A 4solenoid is to be designed to produce a magnetic field of 0.1 T at its centre. The radius is to be 5 cm
and the length 50 cm, and the available wire can carry a maximum current of 10 A.
i.
How many turns per unit length should the solenoid have?
ii.
What total length of wire is required?
[N = 8 x 103 m-1 ;1.3 x 103 m]
[ 8000 turns/metre; 1250 m]

5
Fig. 2 shows two separate wires forming a circle of radius r with centre at O. Both wires carry
equal currents i. Find out the total magnetic field at the centre O due to the two currents.

6
Two parallel wires carry equal currents in opposite directions. If the force per unit length on one of
these wires is 2.5 x 10-3 Nm-1with the wires 20 cm apart, find the current in each wire. Will the wires
attract or repel?
[50 A]

7
The long straight wire AB in the figure shown carries a current of 20 A. The rectangular loop
whose long edges are parallel to the wire carries a current of 10 A. Find the magnitude and direction of
the resultant force exerted on the loop by the magnetic field of the wire.
[7.2 x 10-4 N]



8
What torque is needed to hold a coil of single turn of area 5 cm 2in equilibrium when it carries a
current of 10 mA and is placed in a uniform magnetic field of flux density 0.25 T
i.
with its axis perpendicular to the field,
ii.
with its plane perpendicular to the field,
iii.
with its plane making an angle of 600 with the field?
[1.25 x 10-6 Nm; 0; 6.25 x 10-7 Nm]
9
Two galvanometers are identical in all respects but have coils of 50 and 100 turns which are of
resistance 10 and 600 respectively. What is the ratio of their deflections
i.
when they are connected in series across the same cell?
ii.
when they are connected in parallel across the same cell?
iii.
when they are connected successively to a cell of EMF 2.5 V and internal resistance 2
[1:2; 30:1; 301:12]
1
A horizontal rod PQ of mass 10 g and length 10 cm is placed on a smooth plane inclined at 60 0 to
the horizontal. A uniform vertical magnetic field B acts downwards in the region of PQ as shown. When
the rod carries a current of (3)1/2 A it remains stationary on the plane.
i.
Find the value of the flux density B.
ii.
Find the direction of current in the rod.
[1 Tesla]
Moving Charges in Magnetic Field
Moving charges experience force in magnetic field
If there is already a magnetic field in a region, a stationary charge located in the region would
experience no force. However, if the charge moves, then it sets up its own magnetic field lines. There is an
interaction between the two sets of flux creating a force on the moving charge.
Hence, whenever there is a current carrying conductor placed in a magnetic field not caused by itself,
the current carrying conductor will experience a force due to the interaction of its own magnetic field with that
of the external source. This is because of the net flux on one side of the conductor becoming more than that on
the other as shown in the figure above.
Only if the current is parallel to the external field will it experience no force. This is because its own flux
would be entirely perpendicular to the external flux, making the resultant flux on either side of the current
equal.
If a particle carrying charge Q moving with a velocity v enters a region of uniform magnetic flux
density B directed perpendicular to its motion, then the moving charge is found to experience a force of
magnetic origin, whose magnitude is proportional to the strength of the external field, to the velocity and to the
charge. The magnitude of this force is therefore given by:
In the event that the velocity
of the particle is makes an angle with the magnetic field, then there is
a component ( cos ) of the velocity which is parallel to the flux, and a component ( sin ) which is actually
the velocity v perpendicular to the flux given above. Hence due to the effective motion parallel to the flux there
is no force experienced by the charge, but the force F is due to the perpendicular motion. Hence, the force is
given in magnitude and direction given by the cross product of Qv and B
There would be no force acting on a charge in a magnetic field only if:
a.
The charge is stationary (v=0)
b.
The field strength is negligible (B=0)
c.
The charge moves parallel to the field (sin =0)
Unit of Flux density B
Unit flux density is defined as that in which a particle of charge 1 Coulomb moving at velocity of 1 m/s
normally
to
the
flux
will
experience
a
force
of
1
newton.
This
is
called
1 Tesla (=1 wb/m2 )
Hence, if Q = 1 coulomb, v = 1metre/second perpendicular to the flux density, and if the magnetic force
experienced is 1 newton, then that magnetic field is said to be of unit flux density (1 weber/metre2 or 1 Tesla)
Hence, the constant k is made equal to 1 unit by suitable choice of unit of flux density. Therefore,
Charge moving normally to the field describes circular path at fixed speed
This force F which acts always perpendicular to the velocity of the charged particle does not change the
magnitude of the velocity but does continuously change the direction of motion as shown in the diagram below.
Hence this is a centripetal force constraining the particle to move in a circular path as shown. By
subtracting the initial velocity VA from the final velocity VB and dividing by the time taken the acceleration
a = (VB - VA )/ t can be shown to be given by
The net force Bqv acting on the charged particle of mass m must then be equal to the product of the
mass m and the acceleration a of the charged particle. Hence the equation of the moving particle is given by
Thus, in a particular fixed value of flux density B, a particle of fixed mass m carrying charge Q, the
radius r of the circular path traced by the moving charge is directly proportional to the speed of entry v.
where T is the time period of the motion.
Thus the time period of motion of a particular charge in a magnetic field is constant and independent of
its speed or the radius of its path. Its value depends only upon the mass and charge of the moving particle and
on the flux density of the magnetic field.
Comparison of effects of electric field and magnetic field on charges
CONDITION

charge Q
Stationary
ELECTRIC FIELD
MAGNETIC FIELD

Force = EQ directed

Force = 0

Trajectory -
straight

No

Kinetic
energy
parallel to E
line
change
in
kinetic
energy
changes

Charge
Q

Force of magnitude

moving at speed v parallel EQ directed parallel to E

to the field

Trajectory - straight

line
energy

Kinetic
energy
changes
Force = 0
Trajectory - straight line.
No change in kinetic

Charge
Q

Force of magnitude

Force of magnitude F=Bqv
moving
at
speed
v EQ directed parallel to E
direction normal to B and v
perpendicular to the field

Trajectory parabolic

Trajectory circular with

Kinetic
energy axis parallel to B
changes

No change in kinetic
energy

Charge
Q

Force of magnitude

Force
of
magnitude
moving at speed v at an EQ directed parallel to E
F=BQv direction normal to B and v

Trajectory parabolic

Trajectory - helical (spiral)
angle to the field

Kinetic
energy with axis parallel to B.
changes

No change in kinetic
energy.
Force on current carrying conductor in external magnetic field
Just as a moving charge experiences a force in an external magnetic field, so a stream of moving
charges in a wire also experience similar forces, thus making the whole wire experience equal and parallel
forces along its length when placed in a uniform field. This force is given by a modified form of the same
expression ( F = Qv x B ). Since,
where is the angle between the current vector I l and the field vector B
The magnitude of the force F is given by the product B I l sin and its direction is given by the vector
multiplication rule.
It should be noted that according to change of the angle (from 0o to 90o ) between the current vector
and the flux vector the magnitude of the force changes but its direction remains always perpendicular to the
plane containing the vectors I l and B.
Magnetic forces between two currents
When one current carrying conductor is located within the vicinity of another one, each conductor
experiences a force being in the magnetic field set up by the other. If the two conductors are parallel and
carrying currents in the same direction then the resulting forces pull the two conductors towards each other. If
the currents are antiparallel, then the forces are repulsive.
The Ampere as a fundamental unit.
The flux density at a distance a in air from an infinitely long straight conductor current carrying
current I is given by
This flux being directed perpendicular to the paper, the force on the current carrying conductor A is
given by
The fundamental unit of current in the rationalized MKS unit system is defined from this force of
interaction between the magnetic fields of the two currents thus:
If two infinitely long straight parallel conductors of negligible cross sectional area, carrying equal
currents and separated by a distance of 1.0 m in vacuum, each experiences a force of 2.0 x 10 -7 newton per
metre of length then each current is one ampere.
Permeability of vacuum
Thus, if a = 1.0 metre, and F/l = 2.0 x 10-7 Nm-1 , then I = I' = 1.0 Ampere
Therefore,
Since the units of B are:
the units of 0 are:
weber/m2,
weber/m2 x m/A
=wbA-1 m-1 (= henry/metre)
Such a force acting on a current carrying conductor causes the conductor to move in the direction of
the force it experiences.
Current Carrying coils in Magnetic Field
Forces acting on coil in uniform field
If a rectangular coil carrying a current is placed in a uniform external magnetic field, each arm of the coil
experiences a force according to its orientation with respect to the magnetic field lines.
Let there be a rectangular coil of length l and breadth b having N turns and carrying current I, placed in
a uniform magnetic field of flux density B. Let be the angle between the plane of the coil and the direction of
the field lines. Let be the angle between the normal to the coil and the direction of flux.
The pair of opposite sides PQ and RS bearing current perpendicular to the magnetic field, each
experiences a force equal to B I l N sin 90o. The currents being in opposite directions, the two forces are
oppositely inclined as shown in the figure [b]. These two forces are laterally displaced from each other and thus
constitute a couple which rotates the coil with angular acceleration.
The pair of sides PS and QR (see figure above) carrying currents at an angle to the magnetic field
lines, each experiences a force = B Il N sin . Since this pair of opposite sides carry currents in opposite
directions, the directions of the resulting force on them are also opposite. However, these two equal and
opposite forces have no lateral displacement and are therefore of net torque = 0. These two forces (not shown
in the diagram) are balanced forces. Their combined effect produces neither linear nor angular acceleration,
but tends to produce distortion.
Torque acting on coil in uniform magnetic field
The coil experiences a torque given by
= B I l N x b cos
= B IlN x b sin
= BINA sin
where is the angle between the normal to the coil and the magnetic field and A (= l x b) is the area of
the coil.
As discussed in the chapter on magnetic effect of currents, the coil itself is equivalent to a magnet. Its
clockwise current carrying face is its south pole and the counterclockwise current carrying face is its north pole.
Hence the coil is equivalent to a magnetic dipole perpendicular to the plane of the coil. is thus the angle
between the field lines of the external magnetic field and the magnetic dipole equivalent of the coil.
Dipole Moment
If the coil is located in a uniform magnetic field of unit flux density and held parallel to it (i..e. with axis
perpendicular to the flux density) then it experiences a torque given by
= I N A sin 90o
=INA
This is called the dipole moment of the coil.
Dipole moment is defined as the torque required to hold the dipole perpendicular to a uniform magnetic
field of unit strength.
Units of dipole moment are : Am2
It is a vector quantity, directed from the south pole to the north pole of the dipole
Coil oscillates in uniform field
From
= B INA sin
When =0 the coil rests in equilibrium
If the coil is turned from this rest position in a clockwise direction, it experiences a counterclockwise
torque which turns the coil in a direction opposite to its original deflection. If the initial displacement is
counterclockwise, then the resulting torque is clockwise. That is, it is a restoring torque which causes
oscillation about the mean position.
Thus the general expression for torque is
= - BP sin
The negative sign implies that the displacement and torque cannot both be in the same direction.
This torque causes angular acceleration given by
where
is the angle between the dipole axis and the magnetic field
Oscillation is simple harmonic if initial displacement is small
For very small values of
Thus we have
, sin
=
Since the angular acceleration of the dipole is directly proportional and oppositely directed to its angular
displacement, The dipole performs angular harmonic motion.
Its time period can be found as follows:-
Work is done in turning the coil in the field
If a coil is placed in a uniform magnetic field, it experiences a variable torque which makes it oscillate
about its rest position of alignment with the field. When the initial potential energy is dissipated, it comes to rest
with the coil perpendicular (i.e. with the its magnetic dipole parallel) to the direction of the magnetic field. In
order to rotate the coil from this rest position of lowest potential energy, work has to be done by an external
torque. The work done by an external agent in rotating the dipole through an angle from its rest position is
given by
If the coil is rotated through 180o then work done is 2PB.
If the coil is rotated through 360o then work done is 0
If the dipole is rotated through 90o then work done is PB.
Potential energy of dipole in field
In addition to the work done by an external agent in turning the dipole as calculated above there is also
work done by an external agent in bringing the dipole from infinity to its position in the uniform magnetic field.
The sum total of these two amounts of work done is the potential energy of the dipole in the magnetic field.
The dipole placed at an infinite distance from the magnetic field has zero energy.
Work done is zero if displacement is perpendicular to the force i.e. perpendicular to the magnetic field.
Hence, if the dipole is brought from infinity and placed perpendicular to the magnetic field, then the work done
is zero. The potential energy of the dipole must be zero in this position.
But the dipole left in this position rotates under the system's torque into the position parallel to the
magnetic field. Hence the potential energy is lowest in this position parallel to the magnetic field.
Since ithe potential energy is zero in the perpendicular position and lower than that in the parallel
position, the potential energy must be negative in the parallel position.
In turning the dipole through 90o from its rest position the external torque does work given by
W = PB(1 - cos 90o) = PB
The potential energy in the parallel position is thus lower than that in the perpendicular position by this
amount of work done.
Since potential energy is zero in the perpendicular position, in the parallel position the potential energy
is -PB.
Hence potential energy in any position at an angle with the position of rest is given by
P.E. = - PB + PB(1 - cos ) = - PB cos
Radial Field
The pole pieces of a magnet face each other and they are cut into arcs of a circle as shown below. The
magnetic lines of force are perpendicular to the pole faces but as they can never intersect, they curve away
from each other nearer the centre of the field. If a coaxial soft iron cylinder is placed between the cylindrical
pole faces, then the lines of force are perpendicular to the surface of the soft iron cylinder as well. This makes
the lines of force essentially radial in the gap between the pole faces and the coaxial soft iron cylinder. The
lines of force curve away from their radial path only inside the soft iron cylinder.
Torque on coil in radial field
When the rectangular coil is placed coaxially in a radial field, each of the arms perpendicular to the field
lines experiences a force of
F = B I l N sin
where is the angle between the field lines lying in the plane of the diagram and the current I which is
perpendicular to the plane of the diagram.
Hence two equal forces (given by F = B I / N) act in opposite directions on the two opposite arms of the
coil which are perpendicular to the magnetic field lines. These two oppositely directed forces constitute a
couple rotating the coil about an axis coinciding with that of the magnetic field. As the coil turns these two arms
remain always perpendicular to the magnetic field and the force on each arm at all positions of the coil is:
F = B I/ N
As the coil turns, the perpendicular arms move into a differently directed magnetic field. As shown in
the diagram, at every position of the coil the magnetic field is directed along the breadth b of the coil. The
direction of the force given above is always perpendicular to the plane defined by the current vector I l and the
magnetic field vector B. This plane is therefore the plane of the coil. The perpendicular arm of the couple is
thus always the breadth b of the coil.
Thus the moment of the couple is
= B I /l N b
or,
=BIAN
Since the axis of rotation of the coil passes through the mid point of the arm b and coincides with the
axis of the radial field and of the cylindrical soft iron core, the magnitude of the radial flux density B remains
always a constant value at all positions into which the perpendicular arms move.
The torque acting on the coil has therefore a constant value.
Moving Coil Galvanometer
Between the curved cylindrical pole pieces of a horseshoe magnet is placed a coaxial cylindrical soft
iron core to strengthen the field and render it absolutely radial in the air gap between the core and the pole
faces. A rectangular coil of several turns is also located between the pole faces so that its opposite arms lie in
the air gap between the core and the pole faces. The axis of the radial magnetic field lies along the straight line
joining the midpoints of its other pair of opposite arms. When this coil is connected to the closed circuit in which
the current is to be detected by the galvanometer, it experiences a constant deflecting torque given by
=BIAN
In the simplest form of the instrument, this coil rests on an agate knife edge and is attached to one end
of a spring whose other end is fixed to a rigid support. As the deflecting torque turns the coil, the spring gets
twisted and sets up a restoring torque which is proportional to the angle of twist. The restoring torque therefore
grows until it is equal to the forward torque. About this position the coil oscillates under its own angular inertia
until the dissipation of energy finally brings it to rest. At the rest position,
BIAN=k
where k is the torsional constant of the suspension (i.e. the restoring torque per unit twist)
Thus,
Since the flux density B, the area A of the coil and its number of turns N are also constants, the
deflection is proportional to the current. i.e. the instrument has a linear scale. This is due to the radial field.
The deflection of the coil for each unit current flowing in it is called the current sensitivity of the instrument and
is given by
The strong field magnet serves the dual purpose of rendering the earth's magnetic field negligibly small,
and increasing the current sensitivity of the galvanometer. The current sensitivity of the instrument is also
enhanced by a coil of large number of turns of large area and a spring of low torsional constant (k).
The cylindrical pole pieces create a radial field in which the deflecting couple is of constant torque
(BIAN) which gives rise to a linear scale (I
) for the instrument.
The soft iron core coaxially placed with the coil and the field magnet strengthens the magnetic field and
further ensures that it is indeed radial in the airspace between the core and the pole faces. Its weight is
supported by a brass pin fixed to the body of the instrument as shown below.
The rigid light frame on which the coil is wound prevents the coil from being distorted by balanced
forces acting on the other pair of arms. If the frame is made of metal then as the coil swings in the field, the
frame cutting through flux induces an alternating current in the frame itself. These currents called eddy currents
dissipate the mechanical energy of oscillation and bring the coil rapidly to rest.
The spring sets up the restoring torque which opposes the deflecting torque. It carries the pointer which
moves over a scale to indicate the deflection of the coil. It further serves to complete the electrical circuit.
However, in the spring type of instrument the torsional constant is rather high reducing current sensitivity at the
expense of making the instrument robust.
Alternative to the spring type is the suspension type of moving coil galvanometer in which the restoring
torque is set up by the suspension wire made of phosphor bronze. This wire completes the circuit, supports the
weight of the coil and carries a small mirror just above the coil. A beam of light reflects off this mirror and is
made incident on a long metre scale. The mirror lamp and scale together constitute an optical lever
arrangement which greatly enhance the sensitivity of the instrument due to the low torsional constant (k) of the
wire. However this type of instrument is easily damaged by excess current.