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1. A long copper wire of cross-sectional radius R carries a uniform current I. What is the current density j(r)? Use Ampere's Law to determine B as a function of the distance a from the centre of the wire. Sketch the result.

The current density, the current divided by the area of wire it flows through, is



For cases of radial symmetry, Ampère’s Law

ò_C B•dl = μ₀I_enclosed ,

reduces to

2πaB = μ₀I_enclosed ,          (1)

where r is radius of the Amperian loop centred on the wire. We need only find I_enclosed. To do so we need to consider the two regions here. We will assume that the current is out of the page. Using the right hand rule, this implies that B will circulate counterclockwise (that is B is tangential to the amperian circle in this direction).

1. a < R



The amount of current passing through the amperian circle is proportional to the area

I_enclosed = ò_A j(r) dA = ò₀^a ò₀^2p j(r) rdrdθ = I(πa²/πR²) = I (a²/R²) .

Thus (1) becomes

2πaB = μ₀I (a²/R²) ,

and we find

B = μ₀Ia/2πR² .

2. a > R



The amperian circle encloses the whole wire, so I_enclosed = I.

Thus (1) becomes

2πaB = μ₀I,

and we find

B = μ₀I/2πa .

This is the standard result outside a wire.

A graph of B as a function of R looks like





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2. A long copper pipe with thick walls has an inner radius R and an outer radius 2R. A current I flows along this wall, uniformly distributed over the cross-sectional area of the copper. What is the current density j(r) for all r? Use Ampere's Law to find the magnetic fields as a function of radial distance from the centre of the pipe. Sketch the result.

The current density is the current divided by the area of wire it flows through. Here the area of the pipe is A = π(2R)² – π(R)² = 3πR².



For cases of radial symmetry, Ampère’s Law

ò_C B•dl = μ₀I_enclosed ,

reduces to

2πaB = μ₀I_enclosed ,          (1)

where r is radius of the Amperian loop centred on the wire. We need only find I_enclosed. To do so we need to consider the three regions here. We will assume that the current is out of the page. Using the right hand rule, this implies that B will circulate counterclockwise (that is B is tangential to the amperian circle in this direction).

1. a < R



Here we have no enclosed current, so B is exactly zero.

2. R < a < 2R



The amount of current passing through the amperian circle is given by

I_enclosed = ò_A j(r) dA = ò₀^a ò₀^2p j(r) rdrdθ = I (a² – R²)/(3R²) .

Thus (1) becomes

2πaB = μ₀I (a² – R²)/(3R²),

and we find

B = (μ₀I/2π)(a² – R²)/(3aR²) .

3. a > 2R



The amperian circle encircle the whole wire, so I_enclosed = I.

Thus (1) becomes

2πaB = μ₀I,

and we find

B = μ₀I/2πa ,

which is the standard result outside a wire.

A graph of B as a function of a looks like





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3. A coaxial cable consists of a long cylindrical copper wire of radius r₁ surrounded by a cylindrical insulating shell of outer radius r₂. A final conducting cylindrical shell of outer radius r₃ surrounds the insulating shell. The wire and conducting shell carry equal but opposite currents I uniformly distributed over their volumes. What is the current density j(r) for all r? Find formulas for the magnetic field in each of the regions 0 < a < r₁, r₁ < a < r₂, r₂ < a < r₃, and a > r₃. Sketch the result.

We will assume that the inner cylinder current is out of the page, while the outer shell current is into the page. The current density is the current divided by the area of wire it flows through. Here the area of the inner cylinder is A_inner = π(r₁)². The area of the shell is A_shell = π[(r₃)² – (r₂)²]. Thus we have



For cases of radial symmetry, Ampère’s Law

ò_C B•dl = μ₀I_enclosed ,

reduces to

2πaB = μ₀I_enclosed ,(1)

where a is radius of the Amperian loop centred on the wire. We need only find I_enclosed. To do so we need to consider the four regions here.

1. a < r₁



Since I is out of the page through the amperian circle, this implies that B will circulate counterclockwise (e.g. B is tangential to the amperian circle in this direction). The amount of current passing through the amperian circle is given by

I_enclosed = ò_A j(r) dA = ò₀^a ò₀^2p j(r) rdrdθ = I(πa²/πr₁²) = I (a²/r₁²) .

Thus (1) becomes

2πaB = μ₀I(a²/r₁²) ,

and we find

B = μ₀Ia / 2πr₁² .

2. r₁ < a < r₂



The amperian circle encircles the entire inner cylinder, so I_enclosed = I. Thus

Thus (1) becomes

2πaB = μ₀I

and we find

B = μ₀I/2πa .

3. r₂ < a < r3

   

   The amperian circle surrounds the whole inner cylinder and a portion of the outer shell. The amount of current from the outer shell is proportional to the area enclosed. Hence

   Ienclosed	= òA j(r) dA
   	= ò0a ò02p j(r) rdrdθ
   	= 2πò0a j(r) rdr
   	= 2π{ ò0r1 j(r) rdr + òr1r2 j(r) rdr + òr2a j(r) rdr }
   	= I + 0 – I[(a2 – r22)/(r32 – r22)]
   	= I – I[(a2 – r22)/(r32 – r22)]

   Thus (1) becomes

   2πaB = μ₀I[1 – (a² – r₂²)/(r₃² – r₂²)] ,

   and we find

   B = (μ₀I/2π)[1 – (a² – r₂²)/(r₃² – r₂²)]/a .

4. a > r₃



Here the amperian circle encloses the entire wire. Thus I_enclosed = I – I = 0. Thus the field outside the wire is zero. This is why coaxial cables are popular. They do not produce magnetic interference.

A graph of B as a function of R looks like





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4. A long copper wire of cross-sectional radius R carries a current density j(r) = Ae^-Kr. Use Ampere's Law to determine B as a function of the distance a from the centre of the wire. Sketch the result. The integral identity

   

may be of use.

For cases of radial symmetry and non-constant j(r), Ampère’s Law

ò_C B•dl = μ₀I_enclosed ,

reduces to

2πaB = 2πμ₀ ò₀^a j(r)rdr ,          (1)

where a is radius of the Amperian loop centred on the wire. We need only find solve the integral for the two regions here.

1. 0 < a < R



Equation (1) reduces to

B	= (μ0/a)ò0a j(r)rdr
	= (μ0A/a)ò0a e-Krrdr
	= (μ0A/a){-(r/K)e-Kr + (1/K) e-Kr |0a }
	= (μ0A/a)(1 – e-Ka – Kae-Ka)/K2

2. a > R



Note that j(r) is zero for a > R, so equation (1) reduces to

B	= (μ0/a)ò0R j(r)rdr
	= (μ0A/a)ò0R e-Krrdr
	= (μ0A/a){-(r/K)e-Kr + (1/K) e-Kr |0R }
	= (μ0A/a)(1 – e-KR – KRe-KR)/K2

Despite what its strangeness, this result is actually the standard result for the field outside a wire

B = μ₀I/2πa ,

where I = 2πA(1 – e^-KR – KRe^-KR)/K² .

A graph of B as a function of R looks like





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5. For the following magnetic fields, determine the current density J.

1. B_x = 2xy, B_y = 2z+3, and B_z = 5–2yz.
2. B_x = y³–z³, B_y = z³–x³, and B_z = x³–y³ .

The current density is defined as J = (1/μ₀) Ñ ´B.

(a)

(b)



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6. For the magnetic fields in question 5, determine the current I inside the rectangle with the vertices (4,5,2), (8,5,2), (8,10,2), and (4,10,2). Integrate around the rectangle in the order of the given vertices.

The closed path C described above looks like



The current through C is given by Ampère's Law

I_inside = (1/μ₀) ò_C B·dl           (1)

where we must do the line integral along the given path C. Along segment 1 of path C, dl = idx. Along 2, dl = jdy. Along 3, dl = -idx. And along 4, dl = -jdy. Since B = i B_x + j B_y + k B_z, the integral in equation (1) breaks into four parts

m₀I_inside = ò₁ B_xdx + ò₂ B_ydy – ò₃ B_xdx – ò₄ B_ydy .

Note that along segments 1 y = 5 and z = 2. Along segment 2, x = 8 and y = 2. Along segment 3, y =10 and z = 2. And along segment 4, x = 4 and z = 2.

Case (a):

μ0Iinside	= ò(4,5,2)(8,5,2) 2xy dx + ò(8,5,2)(8,10,2) (2z + 3) dy –    ò (8,10,2)(4,10,2) 2xy dx – ò(4,10,2)(4,5,2) (2z + 3) dy
	= x2y |(4,5,2)(8,5,2) + (2z + 3)y |(8,5,2)(8,10,2) – x2y |(8,10,2)(4,10,2) –    (2z + 3)y|(4,10,2)(4,5,2)
	= (64 – 16)5 + (4 + 3)(10 – 5) – (16 – 64)10 – (4 + 3)(5 – 10)
	= –240

So I_inside = –240/μ₀ Amperes. The minus sign indicates the current is into the rectangle C.

Case (b):

μ0Iinside	= ò(4,5,2)(8,5,2) (y3 – z3) dx + ò(8,5,2)(8,10,2) (z3 – x3) dy –      ò (8,10,2)(4,10,2) (y3 – z3) dx – ò(4,10,2)(4,5,2) (z3 – x3) dy
	= (y3 – z3)x |(4,5,2)(8,5,2) + (z3 – x3)y |(8,5,2)(8,10,2) – (y3 – z3)x |(8,10,2)(4,10,2) –    (z3 – x3)y |(4,10,2)(4,5,2)
	= (53 – 23)(8 – 4) + (23 – 83)(10 – 5) – (53 – 23)(4 – 8) – (23 – 83)(5 – 10)
	= –5740

So I_inside = –5740/μ₀ Amperes. The minus sign indicates the current is into the rectangle C.



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7. A magnetic dipole is given by m = 4i + 7j – 4k. Determine the force on this dipole if it is placed in each of the magnetic fields in question 6.

The force on a dipole in a magnetic field is given by F = (m·Ñ )B. It will be much easier to do the force components separately.

Case (a):

Fx	= mx ¶Bx/¶x + my ¶Bx/¶y + mz ¶Bx/¶z
	= 4 ¶(2xy)/¶x + 7 ¶(2xy)/¶y – 4 ¶(2xy)/¶z
	= 8y + 14x

Fy	= mx ¶By/¶x + my ¶By/¶y + my ¶Bx/¶z
	= 4 ¶(2z + 3)/¶x + 7 ¶(2z + 3)/¶y – 4 ¶(2z + 3)/¶z
	= –8

Fz	= mx ¶Bz/¶x + my ¶Bz/¶y + my ¶Bz/¶z
	= 4 ¶(5 – 2yz)/¶x + 7 ¶(5 – 2yz)/¶y – 4 ¶(5 – 2yz)/¶z
	= –14z + 8y

So the force on the dipole in this case is

F = i (14x + 8y) – j (8) + k (8y – 14z) .

Case (b):

Fx	= mx ¶Bx/¶x + my ¶Bx/¶y + mz ¶Bx/¶z
	= 4 ¶(y3 – z3)/¶x + 7 ¶(y3 – z3)/¶y – 4 ¶(y3 – z3)/¶z
	= 21y2 + 12z2

Fy	= mx ¶By/¶x + my ¶By/¶y + my ¶Bx/¶z
	= 4 ¶(z3 – x3)/¶x + 7 ¶(z3 – x3)/¶y – 4 ¶(z3 – x3)/¶z
	= –12x2 – 12 z2

Fz	= mx ¶Bz/¶x + my ¶Bz/¶y + my ¶Bz/¶z
	= 4 ¶(x3 – y3)/¶x + 7 ¶(x3 – y3)/¶y – 4 ¶(x3 – y3)/¶z
	= 12x2 – 21y2

So the force on the dipole in this case is

F = i (21y² + 12z²) – j (12x² + 12z²) + k (12x² – 21y²) .



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8. In the Bohr model of the hydrogen atom, an electron in the ground state has a speed of 2.20 ´ 10⁶ m/s at a radius of 5.29 ´ 10^-11 m. The charge of an electron is 1.60 ´ 10^-19 C. Find the magnetic dipole moment of the atom.

The dipole moment is defined μ = IA, where I is the current and A is the area enclosed by the current. In the Bohr model an electron is travelling about the core in a circle. Thus A = πR², where R is the radius of the electron orbit. A atomic electron current is given by

I = Δ q/Δ t = e/T ,

where T is the time it takes for the electron to make one orbit. For uniform circular motion, T = 2πR/v. So I = ev/2πR .

Thus the dipole moment is

μ	= (ev/2πR)(πR2)
	= ½evR
	= ½(1.602´ 10-19 C)(2.20´ 106 m/s)(5.29 ´ 10-11 m/s)
	= 9.3 ´ 10-24 Cm2/s

Questions?mike.coombes@kwantlen.ca