2. Magnetic Material
Magnetic field in materials is due to tiny electric
current loops or magnetic dipoles. At normal
temperatures the magnetic moments of the
dipoles are randomly oriented so that the net
magnetic moment is zero.
Magnetization = Magnetic moment per unit
volume
When an external magnetic field is applied the
dipoles align and the material develops a
magnetization.
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3. Dia-, Para-, and Ferronmagnetism
Electric dipoles always align in the direction of
magnetic field but different magnetic material
behave differently.
Diamagnetic material have their manetization
direction opposite to that of B.
Paramagnetic material have M parallel to B
Ferromagnets are those which retain their
magnetization even after the magnetic field is
withdrawn – Hysteresis.
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4. Types of magnetic materials
M = χ m H where χ m is the magnetic susceptibility
If χ mis negative, the material is diamagnetic.
If it is positive, it is paramagnetic.
B = µ0 ( H + χ m H) = µ0 (1 + χ m )H = µ0 µr H = µH
Here μ is the absolute permeability and μ r is the
relative permeability.
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5. Paramagnetic material in Magnetic Field
Atomic magnets are oriented in direction
of B, currents in adjacent loops cancel
giving rise to a surface current
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6. Force on a magnetic dipole
Force on a current carrying loop is
B
L
A
F = I (dl × B)
Hence the force on a closed current loop is zero.
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7. A current loop experiences a force in an
inhomogeneous Field
z
On sides OA & BC
C B
ε ε
F = I ∫ ˆ × B (0, y,0)dy −I ∫ ˆ × B(0, y, ε )dy
j j
0 0
ε
ˆ × ε ∂B
= −I ∫ j dy y
0
∂z 0, y ,ε O A
x
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8. Dipole in Inhomogeneous Magnetic Field
z
On sides AB & OC
ε ε
C B
F = I∫kˆ × B(0, ε , z ) dz −I k × B (0,0, z )dz
ˆ
0
∫ 0
ε
ˆ ∂B
= −I ∫ k × ε dz
∂y 0, 0, z
0
y
O A
x
10/21/12 8
9. Dipole in Inhomogeneous Magnetic Field
Assume that the derivatives are constant at the
boundaries of small loops
ε ∂B
ε
∂B
F = I ∫ (k ˆ ×ε )dz − ∫ ( j
ˆ×ε )dz
0 ∂y 0, 0, z ∂z 0, y ,ε
0
∂B ˆ ∂B y ˆ ∂B ˆ ∂B
= Iε 2 (+ ˆ x − i
j ) − (−k x + i z )
∂y ∂y ∂z ∂z
2 ˆ ∂Bx ˆ ∂Bx + k ∂Bx = ∇(m ⋅ B)
ˆ
= Iε i +j
∂y ∂y ∂z
as ∇ ⋅ B = 0 and m = Iε 2i ˆ
10/21/12 9
10. Torque on a current loop
A dipole experiences a torque in a constant
magnetic field τ = m × B
Torque is zero when the magnetic moment is
parallel to the field. To bring the dipole to a
position which makes an angle θ with the
magnetic field, one has to do work.
θ
W = ∫ τdθ = ∫ mB sin θdθ
0
= mB(1 − cos θ )
Potential Energy of dipole
U = -m ⋅ B
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11. Field due to a magnetic dipole
For paramagnetic material, magnetization is
proportional to the external field.
Magnetic field on the axis of a circular coil of radius
a was seen to be
µ0 I a2 µ m
⇒ 0 3 (for z >> a )
2 (a 2 + z 2 ) 3 / 2 2π z
very similar to that of electric field for an
electric dipole !
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13. Vector Potential of a current loop
z
current loop is in x - y plane x
P
field point in x - z plane dA
r
ρ = distance of P from d l ρ
θ
r = position relative to O α
x O
φ
dl
y
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14. z
x
P
dA
since dA|| dl , A has no z - component.
r
For each d l , ∃ a symmetrically opposite ρ θ
dl’
element which cancels x - component of
α
dA but adds y - component of dA x O
φ
dl
y
10/21/12 14
15. Vector Potential of a current loop
µ 0 I dl µ 0 I 2π a cos ϕ
ˆ
A=
4π ∫ ρ 4π 0 ρ dϕ
= k∫
µ 0 I ˆ 2π a cos ϕ
= k∫ 2 dϕ
4π 0 (r + a − 2ar cos α )
2 1/ 2
µ 0 I ˆ 2π 1 a 2 2a cos α −1/ 2
≈ k ∫ a cos ϕ (1 + 2 − ) dϕ
4π 0 r r r
µ 0 I ˆ 2π 1 a 2 a cos α 3a 2 cos 2 α
= k ∫ a cos ϕ (1 − 2 + + ) dϕ
4π 0 r 2r r 2r 2
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16. Vector Potential of a current loop
Use r cos α ≈ x cos ϕ
µ 0 I 2π 1 a 2 ax cos ϕ 3a 2 x 2 cos 2 ϕ
A= ˆ
k ∫ a cos ϕ (1 − 2 + + ) dϕ
4π 0 r 2r r 2
2r 4
µ 0 I ˆ 2π ax
= k ∫ a cos ϕ 3 dϕ
2
4π 0 r
µ 0 Ia 2 ˆ x µ mx ˆ µ m sin θ ˆ
= k 3π = 0 3 k= 0 2 k
4π r 4πr 4πr
µ0
= m×r
4πr 3
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17. Vector Potential of magnetized substance
If M is the magnetization, a volume dτ contains
Mdτ dipoles. The vector potential at a point P is
given by
µ 0 M × (r − r ′)
A(r ) =
4π ∫ r − r ′ 3 dτ ′
µ0 1
=
4π ∫ M × ∇′ r − r ′ dτ ′
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18. Vector Potential of magnetized substance
Use
M 1 1
∇′ × = ∇′ × M − M × ∇′
r − r′ r − r′ r − r′
µ0 1 1
′ × Mdτ ′ − ∫ M × ∇′ dτ ′
4π ∫ r − r ′
A(r ) = ∇ r − r′
µ0 1 1
4π ∫ r − r ′
= ∇′ × Mdτ ′ + ∫ M × dS ′
r − r′
In the last step, we have used
∫ ∇ × vdτ = − ∫ v ×ds
vol surface
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19. Bound currents
µ0 1 1
A( r ) = ∫ ∇′ × Mdτ ′ + ∫ M × dS ′
4π r − r ′
r − r′
µ 0 J b (r ′) µ0 K b (r ′)
=
4π vol∫ r − r ′ dτ ′ + 4π surface r − r ′ dS
∫
J b (r ) = ∇ × M
K b (r ) = M × n ˆ
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21. Free and Bound Currents
Net current is due to actual transport of
charges while bound current is due to
magnetization.
J = J free + J bound
Ampere’s Law for magnetized material
contains contribution due to both types of
current and is,
∇ × B = µ0 J = µ0 ( J f + J b )
= µ0 ( J f + ∇ × M )
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22. Fields B, H and M
B
Define H-field by H = −M
µ0
∇× H = J f
Equivalently,
∫ H ⋅ d l =I free(enclosed )
Also
10/21/12 ∇ ⋅ H = −∇ ⋅ M 22
23. Example
An infinitely long cylinder of radius R contains a
frozen in magnetization M=kr, where r is distance
from axis and no free current. Find B, H, M.
M = krk ˆ
Bound volume current
1 ∂M z ∂M z
Jb = ∇ × M = r−
ˆ ϕˆ
r ∂ϕ ∂r
= −kϕ ˆ
ˆ ˆ ˆ
K b = M × n |r = R = krk × r |r = R = kRϕˆ
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24. Infinitely long cylinder
Use Ampere’s Law. As the current is
in φ direction, field are along the axis. B is zero
outside the cylinder.
R
∫ B ⋅ dl = − BL = µ0 (− K b L + ∫ J b Ldr )
r
= µ 0 (−kLR + (+ k ) L( R − r )) = − µ 0 kLr r
L
ˆ
B = µ 0 krk
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25. Magnetized cylinder (contd.)
∂M z
∇⋅M = = 0 , M is specified by curl alone.
∂z
B
∫ H ⋅ dl = 0 ⇒ H = 0 = µ 0 - M
B = µ 0 M = 0 outside cylinder
ˆ
= µ kr k
0
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26. A uniformly magnetized Cylinder
z
Jb = ∇ × M = 0
θ dl × r
ˆ K b = M × n = M sin θ
ˆ
M
R
θ
Rsin θ
n
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27. Uniformly magnetized sphere
From symmetrically located pairs of elements horizontal
components cancel while vertical components add up.
µ0 dl × r
ˆ µ dl sin θ ˆ
dB = dI 2 ⇒ 0 dI k z
4π r 2π R2
dI = K b Rdθ = MR sin θdθ
contribution to dB from the ring
θ dl × r
ˆ
µ0 sin 2 θ
dB ring = MR dθ ∑ dl k ˆ
2π R 2
M
µ0 sin 2 θ ˆ µ ˆ R
= M 2πR sin θk = 0 M sin 3 θdθk θ
2π R 2
µ0 ˆπ 3 µ ˆ 4
B= Mk ∫ sin θdθ = 0 Mk × Rsin θ
2 0
2 3 n
2
= µ0 M
3
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28. Example : Large Piece of uniformly magnetized
material from which a sphere is scooped out.
For the material without void,
B0
H0 = −M
µ0
Fill the void with a sphere of uniform
magnetization M
2
B′ = B0 − µ 0 M
3
B
H= −M
B0 µ0
Since at the centre of the hole M = 0,
B′ B0 2 H 0 1
H′ = = − M= + M
µ0 µ0 3 µ0 3
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