# PHYS2939 Electromagnetism Solution to Assignment 2 Session 1 2002

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```					                                             PHYS2939 Electromagnetism
Solution to Assignment 2 Session 1 2002

Question 1

(a) We take B due to a single turn (as derived in lectures; see lecture notes Section 2.2.2)

µ 0 I dlxrˆ        µ I                        dlxr ˆ
dB =           2
and B = 0                     ∫      2
4π r                4π                         r

The total magnetic induction is along the axis (symmetry – the components pointing in directions
perpendicular to the loop’s axis sum to zero) and we see that

µ 0 I 2 πa             µ 0 Ia 2                        B on axis of a circular
Bz =              cos θ =               3
4π r 2             2( a 2 + z 2 ) 2                    current loop, radius a

For a solenoid length l with n turns per unit length we sum up the contribution from each loop to obtain

µ0       l 2          Ia 2 dz
Bz =
2    ∫
−l 2
n
(a 2 + z 2 )
3
2

1                    t
The integral   ∫ (1 + t   2 32
)
=
(1 + t 2 )1 2
is standard and we find

nµ 0 a 2 I    z    +l 2
Bz =             2 2          ˆ
z
2  −l 2
2 a a + z 

nµ 0 I    z+l 2         z−l 2        
Bz =           2           − 2                ˆ
 z
2  a + (z + l 2 )2
               a + (z − l 2)2 


Centre of solenoid, on axis: if we put z = 0 ,

nµ 0 I    l                   l2
Bz =           2         ˆ
 z = nµ 0 I 2              ˆ
z

2
2  a + (l 2 )             a + (l / 2 ) 2
If we choose the centre as our reference and measure θC as shown here
we find
ˆ
Bz = nµ 0 IsinθC z

θC        θΕ

End of solenoid, on axis: we put z = ± l 2 and obtain

ˆ
Bz = nµ 0 Isinθ E z

(b) Helmholtz coils

(i) Recall (again) the expression for B on the axis of a single turn coil
(current loop) derived in lectures:
Fig 2.
µ 0 I dlxrˆ        µ I           dlxr ˆ
Bz                   dB =           2
and B = 0        ∫      2                    a
4π r                4π            r                                        z
d

I

(see lecture notes Section 2.2.2 if you’ve forgotten)
The total magnetic induction is along the axis (symmetry – the components pointing in directions
perpendicular to the loop’s axis sum to zero) and we see that

µ 0 I 2 πa             µ 0 Ia 2            B on axis of a circular
Bz =              cos θ =               3
4π r 2             2( a 2 + z 2 ) 2        single current loop, radius a

The Helmholtz geometry is

a
z
d
Taking z = 0 half way between the coils, the above expression for Bz gives us

µ 0 Ia 2     1                                                                1   
B=                                                      32       +                       32 
[
2
2  a + (d 2 + z)2
                                    ]             [   a 2 + (d 2 − z)2         ]

∂B 3µ 0 Ia 2  −( d 2 + z)                                                               (d 2 − z)   
(ii)                                          =                                                           52        +                       52

∂z    2  a 2 + (d 2 + z)2
                 [                              ]               [   a 2 + (d 2 − z)2     ]
and at z = 0
∂B        3µ Ia 2    −( d 2 )                                                         (d 2)     
= 0                                                 52      +                       52
=0
∂z z = 0    2  a + (d 2)2

2
[                    ]                 [   a 2 + (d 2)2  ]


∂ 2 B 3µ 0 Ia 2
=          x
∂z 2     2
      −1               (d 2 + z)( − 5 2)2(d 2 + z)                                            −1                             (d 2 − z)( − 5 2)2(d 2 − z)( −1) 
              52   −                                    72                   +                                  52      +                           72        
[
 a + (d 2)2
2
]                [a   2
+ (d 2)2       ]                         [a   2
+ d 2 − z)2       ]                          [
a 2 + (d 2 − z)2    ]      


∂2 B      3µ Ia 2     −2                               2 ( 5 2 )2 ( d 2 ) 2 2      3µ 0 Ia 2                                       2 d 2 5d 2 
2
= 0                               52   −                        72   =                                              72   − a −   +   
∂z z = 0
     [
2  a 2 + (d 2)2               ]            [a 2 + (d 2)2           a 2 + (d 2)2
  ]            [                              ]           4   4 

3µ 0 Ia 2
=                             2 72
{d   2
− a2}
[a   2
+ (d 2)           ]

when d = a (coils separated by the
∂2 B
∂z 2 z = 0

When the coils are separated by a distance equal to the coil radius (d = a) we find

µ 0 Ia 2    1                                             1                      1     8µ I
Bz = 0 =                                   32      +                        = µ 0 Ia 2           = 3 20
2 32                 2   32
[ 2
2  a + (a 2 ) 2
                    ]              [    2
a + (a 2 )      
      ]     (5a 4)     5 a
(iv) Graph of B(z) for the coil separations indicated looks approximately like:

B(z)
d<a

d=a

d>a

z=0                      z

Is this reasonable? Yes. Each coil, separately has a B profile:

B (z)
z

For a pair of coils, the total field between the coils is the sum of the two individual fields:
It is clear that the total B varies with
coil separation.
Bcoil 1                             Bcoil 2

(v) The Helmholtz coils can provide a uniform B region when d = a. They may therefore be viewed as
the magnetostatic analogue of the parallel plate capacitor (which produces a uniform E field between
plates).

Question 2

(i) The carrier density n (number of conduction electrons per cubic metre).

ρN A        -3
For copper, each atoms supplies one free electron. There are           atoms.m
A
ρ is density of copper, 8.9x10 3 kgm −3 ,
A is atomic weight of copper, 63.3,
NA is Avogradro’s number, 6.02 x10 23 (g.mol)−1 or 6.02 x10 26 ( kg.mol)−1
(8.9x10 3 )(6.02 x10 26 )
∴ n=                         = 8.46x10 28 m −3 for Cu
63.3
For aluminium, each atom supplies three conduction electrons.
ρ for Al is 2.7x10 3 kgm −3 ,
A for Al is 27
(2.7x10 3 )(6.02 x10 26 )
∴ n=3                                 = 1.8x10 29 m −3 for Al
27

V                     L    L
(ii) The current is given by I =              and R track = ρ = ρ
R track                A    W .t

ρ is the resistivity (of Cu or Al ), Semiconductor Roadmap web-page gives
ρCu = 2.2 µΩ.cm = 2.2 x10 −8 Ω.m , for ρAl we multiply this by 1.6 (as given in the assignment question)
so ρCu = 3.74 µΩ.cm = 3.74 x10 −8 Ω.m
L is the track length, L = 10 µm
A is the track cross-sectional area, A = W.t where W is the track width, W = 1µm; t is the track
-9
thickness (cladding thickness). Semiconductor Roadmap web-page gives t = 14 nm = 14x10 m (2002
figure, barrier/cladding thickness for intermediate Cu wiring – notice on the webpage, this thickness is
expected to be halved by 2007)

−8       10 x10 −6 m                              V       1V
⇒ R track , Cu   = 2.2 x10 .                            = 15.7 Ω      I=          =       = 63.6 mA (Cu track)
(1x10 −6 m )(14 x10 −9 m )                    R track 15.7 Ω

10 x10 −6 m                           V       1V
⇒ R track , Al = 3.74 x10 −8.                              = 26.7 Ω   I=          =       = 37.4 mA (Al track)
(1x10 −6 m )(14 x10 −9 m )                 R track 26.7 Ω

(ii) The drift velocity is found from
I
I = nAv d e ⇒ v d =
nAe
n = number density of carriers, A = conductor cross-section , vd = drift velocity, e = carrier charge (on
electron).

I               63.6x10 −3
Drift velocity in copper track: v d , Cu =       =        28     −6       −9      −19
= 335.6 ms −1
nAe (8.46x10 )(1x10 x14 x10 )(1.6x10 )

I               37.4 x10 −3
Drift velocity in aluminium track: v d , Al =        =       29     −6         −9    −19
= 92.7 ms −1
nAe (1.8x10 )(1x10 x14 x10 )(1.6x10 )
I
(iii) The current density is J =     (amps per metre squared)
A
63.6x10 −3
For copper tracks, J Cu   =      −6        −9
= 4.5x1012 Am −2
(1x10 x14 x10 )

37.4 x10 −3
For aluminium tracks, J Al =                             = 2.67x1012 Am −2
(1x10 −6 x14 x10 −9 )

(iv) The tracks are essentially very thin flat sheets. The field B at one track, say at point P (see
diagram) due to the current flowing in the adjacent track is (see Problem 1, Tutorial Sheet 5),

µ0I      W                                 W
B=         ln 1 + 
2 πW      b
b
For copper,                                                                                    I
−7
( 4 πx10 )(63.6x10 )   −3
1x10           −6            P
B=                 −6
ln1 +     −6 
= 7.6x10 −3 T
2 π(1x10 )       1.5x10 

where b is taken from the edge of one track to the centre of the adjacent track, b = 1.5 µm.

For aluminium, the current is a factor x(37.4 63.6) smaller so B = 4.5x10 −3 T

Recall the Earth’s field is approximately 0.5x10 −4 T so the B calculated above is roughly 10x the
Earth’s field. Thus the calculated B is small but not negligible.

Question 3

(i) The magnetic induction due to a straight wire is

µ0I
B=
2 πr
so the field at wire 2 due to wire 1 is
µ 0 I1
B2 =
2 πr
resulting in a force ( dF = IdlxB )

µ I
F = I 2  0 1  ∫ dl
 2 πa 
giving a magnetic force per unit length
µ 0 I1 I 2
Fm =
2π a
and with I1 = I 2 = λv ,
µ 0 λ2 v 2
Fm =
2π a
The electric field of one line of charge is

1 λ
E=
2 πε 0 s

giving an electric repulsion per unit length of one wire on the other

1 λ2
FE =                                [ E=F/q].
2 πε 0 a

(ii) Obtain an expression for the ratio of the electric and magnetic forces between the wires in terms of
the carrier speed v and velocity of light c.

µ 0 λ2 v 2
Fm =
2π a

1 λ2
FE =
2 πε 0 a
The ratio is
Fm FE = µ 0 ε 0 v 2

1
Recalling that c 2 =          (c is velocity of light) we see
ε 0µ 0

v2
Fm FE =
c2

(iii) Comment on the result of part (ii): The electric and magnetic forces are equal when v = c. At
velocities v << c (the range ‘normally’ encountered) the magnetic force is very weak compared to the
electric force. We cannot accelerate a physical line of charge (e.g. electrons) to the velocity of light as
this would be in violation of one of the postulates of Special Relativity (recall your relativity lectures in
PHYS1231).

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