COULOMB'S LAW OF FORCE FOR MAGNETIC POLES
(i) Like poles repel each other while unlike poles attract each other.
(i) The force between two magnetic poles is Directly proportional to the product of their pole strengths and Inversely proportional to the square of distance between them.
* Consider two poles of magnetic strength m1 and m2 placed at a distance ’ d ‘ apart in a medium
* According to Coulomb's laws, the force between the two poles is given by:
Where K Is a constant whose value depends upon the surrounding medium and the system of units employed.
K = μ0 μr / 4π
Where , μ0 = Absolute permeability of vaccume or air
μr = Relative permeability of surrounding medium
* μr = 1 , for vacuum or air
* μ0 = 4π x 10-7 H /m
So, Magnetic force in air or vacuum is
Define 1 Ampere- meter
* If , m1 = m2 = m ,
d = 1 m , and F= 10-7 N
* then from force formula
so, m= + 1 or -1
hence , If two equal poles are kept in vacuum at 1 m distance and they exert 10-7 N force to each other , then pole strength will be 1 A-m each.
Magnetic Dipole
If two unlike
magnetic poles of equal strength are
separated at small distance, then system of two poles is called magnetic
dipole.
e.g Bar magnet,
A current loop , magnetic needle, solenoid
etc. are magnetic dipoles.
 |
| Magnetic dipole |
* Magnetic strength of north pole is represented by m
* Magnetic strength of south pole is represented by ( - m )
* Distance between two poles of magnet is called magnetic
length ( 2l )
* SI unit of magnetic pole is
Ampere meter ( Am ).
Magnetic dipole moment
It is the product of magnetic strength of either pole
and magnetic length.
i.e Magnetic
Dipole (M ) = m
x 2l
* Magnetic dipole is a vector quantity
* Its direction is from south
pole to north pole
* Its SI unit is
(i) ampere meter2 ( Am2 )
(ii) Joule per tesla ( JT-1 )
Direction of Magnetic field produced by magnetic poles
* Magnetic field produced by north pole at any point is taken away from North pole from that point on line joining pole and point.
* Magnetic field produced by south pole at any point is taken toward south pole from that point on axis on line joining pole and point.
Magnetic flux density a point due to bar magnet
Case (1) When the point is situated on axial line of bar magnet
* Magnetic flux density at point P due to north pole
* Magnetic flux density at point P due to south pole
Net flux density at point P
Baxial =
B1 – B2
If bar magnet
is very small . ie. d >> l
So, l2
is neglected as compared to d2
.
Thus,
Direction of net Magnetic field
(a) When point
is situated on the side of North pole
* Direction of net magnetic flux density at point P is away
from North pole
(b) When point is situated on the side of South pole
* Direction of net magnetic flux density at point P is toward
South pole
Case (2) When the point is situated on equatorial line of bar magnet
* Magnetic flux density at point P due to north pole
* Magnetic flux density at point P due to south pole
Here , Magnitude
of flux density are equal
i.e B1
= B2
* Vertical components of fields cancel each other
* Net flux density at point P = Sum of horizontal component
I.e Beqa = B1 cos θ + B2 cos θ
Or, Beqa =
2 B1
cos θ (
Because , B1 = B2 )
If bar magnet is very small . ie.
d >> l
So, l2
is neglected as compared to d2
.
* Direction of net magnetic flux density at point P is toward
South pole parallel to magnetic axis.
Relationship between Baxial and Beqa
Torque on Magnetic dipole in uniform magnetic field
A dipole when
placed in uniform magnetic field experiences magnetic torque ( T ) .
* When a dipole is placed at angle θ to a uniform magnetic
field , its magnetic poles experience equal and opposite force parallel to
magnetic field.
Thus, forces
acting on poles produces magnetic torque.
* Force on north pole =
mB ( in direction B )
* Force on south pole
= mB ( In opposite direction of B )
* Distance between line of two parallel force = 2l sinθ
Thus, Magnetic
torque = Magnetic force on either pole X
distance between line of forces
i.e T = mB x 2l sinθ
or, T = 2ml
x B sinθ
i.e T =
MB sinθ
* direction of torque is perpendicular
to plane containing M and B vectors.
* Torque is a vector quantity. Its SI
unit is N-m.
Vector Form of Torque formula
Case(1) if θ = 00
Then , T =
MB sin00 = 0 (
Minimum value of torque )
Thus, Tmin
= 0
Case(2)
if θ = 900
Then, T = MB
sin900 = MB
( maximum value of torque )
Thus, Tmax = MB
Magnetic dipole moment in term of torque
We have, T = MB sinθ
If B = 1 T ,
and θ = 900
Then , T = M
* Thus, When a magnetic dipole is placed in uniform magnetic field
of strength 1 T at 900 to direction of field, Then torque exerted on
dipole is magnetic dipole moment .
SI Unit of M : Nm/T , J/T
, Am2
Work Done to rotate magnetic dipole away from equilibrium position
When a dipole is rotated in uniform magnetic field away from equilibrium position then work is done
on system.
* Let a dipole be at angle θ
to direction of magnetic field.
* At this position Torque on dipole = T
* Work done to rotate small angle dθ,
dW = T dθ
So, Work done
to rotate from initial angle θ1 to final angle θ2
Thus, W= MB ( cosθ1
- cosθ2 )
Where, θ1 = Initial angle between dipole moment and field
θ2 = Final angle between dipole moment and field
Potential energy of magnetic dipole in uniform magnetic field
* When a dipole is placed in uniform magnetic field at angle θ from equilibrium position then work is done on
system.
* This work done is stored in system as potential energy.
* Work done to rotate dipole from equilibrium position to
angle θ
W= MB ( 1 - cosθ )
Thus , Potential energy of Dipole in uniform magnetic
field placed at angle θ from equilibrium
U
= MB ( 1 - cosθ )
Or, U = MB – M.B
Current Loop as Magnetic Dipole
* When current flows in
closed loop of conductor , its one face acts like north pole and other face is
south pole .
* Polarity of face of loop is identified by Clock face rule or
ampere’s thumb rule.
* Thus, a current carrying loop behaves as a system of two
equal and opposite magnetic poles and hence is a magnetic dipole.
Magnetic dipole moment expression :
The magnitude of magnetic moment of loop of current
loop depend on
(i) Dipole
moment is directly proportional to
Current in loop
i.e M ∝ I
(ii) Dipole moment is directly proportional to area of
loop
i.e M ∝ A
Combining
relationship,
M ∝ I A
Or, M = K
I A
* The SI unit of magnetic dipole moment of loop is so defined that K = 1
Thus, M = I A
* If there are n turns in loop
Then , M = n I A
* Direction of magnetic moment is perpendicular
to plane of loop and directed to south face to north face of loop .
* SI unit of magnetic moment = Am2
Magnetic dipole moment of solenoid
* In solenoid there are large number of turns
closely wound .
Each turn has dipole moment
= IA
Where , I = current in
solenoid, and A= area of surface of each turn
Let there are n number of turns
in solenoid,
So, Magnetic dipole moment of solenoid
M = n I
A
* Direction of magnetic moment is from south
pole to north pole of solenoid.
AN ATOM AS A MAGNETIC DIPOLE :
* In an atom, electrons revolve around the nucleus. The revolving electron is equivalent to a current loop.
* One face of loop behave as North pole and opposite face is south pole . Thus , current loop behaves as a magnetic dipole.
* Since a current loop behaves as a magnetic dipole, due to this an atom is like a magnetic dipole and possesses definite magnetic dipole moment ( M ) .
* Direction of dipole moment is from south pole to North pole .
Expression for magnetic dipole ( M ) :
* Let An electron revolve around the nucleus with uniform angular velocity ω and radius of orbit is r .
* The revolving electron is equivalent to the single turn current loop.
Therefore magnetic moment M of current loop is given by ,
M = I A ( I = current in loop )
Now, I = e /T , where T is time period of revolving .
And , T = 2π / ω
So, I = e / 2π / ω
i.e I = e ω / 2π
Area of current loop, A = π r2
So, magnetic moment of atom ,
M = e ω / 2π x π r2 = e ω r2 / 2
Or, M = evr / 2 ----- ( I ) ( v = ωr )
* Direction of dipole moment is from south pole to North pole .
According to Bohr’s
Angular momentum of an electron is integral multiple of h / 2π
i.e, mvr = nh / 2π
or, vr = nh / 2π m
( n is positive integer , and h is plank constant )
Put value of vr in eq (1),
M = n eh / 4π m
eh / 4π m is constant for each electron . it is called Bohr magneton ( μB )
μB = eh / 4π m
Thus , dipole moment , M = n μB
Define Bohr Magneton
Dipole moment of an atom due to electron of first orbit is called Bohr magenton.
Bohr Magneton , μB = eh / 4π m
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