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Electricity and Magnetism: PHY-204.
11 November, 2014
Collaborative ASSIGNMENT
Assignment 3: Sources of magnetic fields
Solution
1. (a) A conductor in the shape of a square loop of edge length l m carries a current I
as shown in the figure below. Calculate the magnitude and direction of the magnetic
field at the center of the square. (b)What If ? If this conductor is formed into a single
circular turn and carries the same current, what is the value of the magnetic field at
the center?
Answer
z
dz
y
90+θ
r
x
z
θ
l/2
Field inwards.
Due Date: 18 November, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
11 November, 2014
|dBz | =
µ0 I|d⃗ℓ × r̂|
(
( ))
4π z 2 + ℓ 2
2
=
µ0 I dz sin( π2 + θ)
( ))
(
4π z 2 + ℓ 2
2
µ0 I
=
(
( )2 ) cos θ dz.
4π z 2 + 2ℓ
Now
z
ℓ/2
= tan θ,
|dBz | =
=
∴ |Bz | =
=
=
⇒ z = 2ℓ tan θ ⇒ dz = 2ℓ sec2 θ dθ.
( )2
µ0 I
ℓ
sec2 θ dθ
cos θ
( ℓ )2
2
2
4π 2 (sec θ)
µ0 I
cos θ dθ
2πℓ
( )
∫ + tan−1 z
ℓ/2
µ0 I
( ) cos θ dθ
z
2πℓ θ=− tan−1 ℓ/2
+π/4
µ0 I
sin θ−π/4
2πℓ (
)
µ0 I
1
µ0 I
· √
=√
· (Fields are additive in this case)
1
2πℓ
2 + √2
2πℓ
Furthermore a wire of length 4ℓ implies a circle of radius
2ℓ
.
π
2l
π
Field at its center is,
µ0
B =
I
4π
∫
2π (
ϕ=0
)
2ℓ dϕ
( 2ℓ )2
π
π
µ0 I π
· 2π
=
4π 2ℓ
µ0 πI
=
·
4ℓ
Due Date: 18 November, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
11 November, 2014
2. A conductor consists of a circular loop of radius R and two straight, long sections as
shown in the figure below. The wire lies in the plane of the paper and carries a current
I. Find the vectorial expression for the magnetic field at the center of the loop.
Answer
From infinite wire, B1 =
From circle, B2 =
µ0 I
,
2R
µ0 I
,
2πR
into the plane of the paper.
also into the plane of the paper.
(see the previous question). Hence
B=
µ0 I
µ0 I
+
,
2πR
2R
into the plane of the paper.
3. A sphere of radius R has a uniform volume charge density ρ. Determine the magnetic
field at the center of the sphere when it rotates as a rigid object with angular speed ω
about an axis through its center, as shown on the next page.
Answer
Due Date: 18 November, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
11 November, 2014
(a)
(b)
(c)
Rmax(z)
r
z
Rmax
r
O
Rmax(z)
θmax
θ
Section showing a
disk at height z
We break the sphere into disks. Each disk is at height zϵ[−R, R] from the equatorial
√
plane. Each disk comprises rings at radius rϵ[0, Rmax (z) = R2 − z 2 ] from the center
of the disk.
We first find the current due to a ring. this is given by
i = ω ρ r dr dz.
This ring produces a field at O, whose z-component is given by
µ0 ω µ r3 dr dz
·
2 (z 2 + r2 )3/2
(In class we have derived the axial field due to a current carrying loop). Now the total
field at O due to a disk at height z will be given by,
µ0 ωρ
dz
2
Let’s first solve the integral
∫
√
Rmax = R2 −z 2
r=0
∫ Rmax
r=0
r3 dr
(z 2 +r2 )3/2
= G.
Let z 2 + r2 = x ⇒ 2rdr = dx ⇒ rdr =
Due Date: 18 November, 2014, 5:00 pm
r3 dr
·
(z 2 + r2 )3/2
dx
.
2
r3 dr = (x − z 2 ) dx
. Therefore
2
4
Electricity and Magnetism: PHY-204.
∫
R2
G =
z2
R2
∫
=
z2
1
=
2
[∫
11 November, 2014
(x − z 2 ) dx
x3/2 2
(since Rmax = R2 − z 2 )
x−3/2
(x − z 2 ) dx
2
R2
x
−1/2
z2
z2
dx −
2
∫
]
R2
x
−3/2
dx
z2
2
R
−1/2 R 2
z2 x
1 x1/2 −
=
2 1/2 z2
2 (−1/2) z2
[
]
1
2 1
= (R − z) + z
−
R z
z2
= R−z+
−z
R
z2
= R+
− 2z.
R
Hence the magnetic field due to the entire revolving sphere is:
)
(
∫
µ0 ωρ +R
z2
− 2z dz
R+
2
R
−R
)
(
∫
2µ0 ωρ +R
z2
=
− 2z dz
R+
2
R
0
]
[
R
1 3 R 2 2 R
z
− z 0
= µ0 ωρ Rz 0 +
3R 0
2
[
]
2
R
= µ0 ωρ R2 +
− R2
3
µ0 ωρR2
=
·
3
4. A hollow cone (like a party hat) has vertex angle 2θ, slant height L and surface charge
density σ. It spins around its symmetry axis with angular frequency ω. What is the
magnetic field at the tip?
Answer
Due Date: 18 November, 2014, 5:00 pm
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Electricity and Magnetism: PHY-204.
11 November, 2014
90-θ
θ
l
z
R(l )
dl
Charge on the dashed strip = σ2πR(ℓ) dℓ
Current i due to the strip = σωR(ℓ) dℓ
= σωℓ sin θ dℓ.
Field due to a small section inside the ring
=
µ0 iR(ℓ) dϕ
µ0 iℓ sin θ dϕ
=
·
2
4π
ℓ
4π
ℓ2
Field’s z-component due to a small section
=
µ0 iℓ sin2 θ dϕ
·
4πℓ2
Field’s z-component due to the ring at height ℓ
∫
µ0 iℓ sin2 θ 2π
=
dϕ
4πℓ2
ϕ=0
µ0 iℓ sin2 θ
2ℓ2
µ0 σω sin3 θ dℓ
=
·
2
=
Total field
µ0 σω sin3 θ
B=
2
Due Date: 18 November, 2014, 5:00 pm
∫
L
dℓ =
0
µ0 σω sin3 θL
·
2
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Electricity and Magnetism: PHY-204.
11 November, 2014
5. How should the current density inside a thick cylindrical wire depend on r so that the
magnetic field has constant magnitude inside the wire? I am looking for a mathematical answer.
Answer
R
r
Assuming that the wire has radius R, consider a circular Amperian Path inside the
wire at radius r, where r < R, as shown in the diagram above.
∫
∫
⃗ ⃗l = µ0
⃗
⃗j(⃗r ).dA
B.d
ClosedP ath
ShadedArea
⃗ are parallel, then the RHS becomes
where LHS = B(2πr). Suppose ⃗j(r) and dA
∫
µ0
∫
∫r ∫2π
⃗ = µ0
⃗j(⃗r ).dA
ShadedArea
j(r)dA = µ0
ShadedArea
∫r
j(r′ )r′ dϕdr′
r′ =0 ϕ=0
j(r′ )r′ dr′ .
= µ0 2π
r′ =0
Now B at the radius r is given by
µ0
B=
r
∫r
∫r
j(r′ )r′ dr′ .
r′ =0
If B is independent of r, r′ =0 j(r′ )r′ dr′ must be proportional to r, that is,
∫r
j(r′ )r′ dr′ = Kr, where K is a constant.
r′ =0
∫r
∫r
By inspection we must have j(r′ )r′ = K, such that r′ =0 j(r′ )r′ dr′ = r′ =0 Kdr′ = Kr.
Hence j(r′ ) = K/r′ or j(r) = K/r. Thus the current density must taper off as 1/r
inside the conductor if the field inside the conductor remains uniform.
Due Date: 18 November, 2014, 5:00 pm
7