PHYS 1220 Laws of Biot-Savart & Ampere Solutions

Questions

Laws of Biot-Savart & Ampere Solutions

In diagram A below, two wires are carrying currents, I₁ = 30.0 A and I₂ = 22.0 A, in opposite directions. Diagram B gives a head on perspective. Determine the direction and magnitude of the net magnetic field at points A, B, and C where r₁ = 0.250 m, r₂ = 0.350 m, r₃ = 0.700 m, and r₄ = 0.500 m.

The magnetic field is tangential to the cross-section of a wire. The direction is clockwise or counterclockwise using a Right Hand Rule. To use the Right Hand Rule, you curl your hand around the wire with your thumb pointing in the direction of the current. The magnetic field "circulates" in the same direction that your hand is curling.

Point A:

The magnitude of the magnetic field due to a wire is given by B = μ₀I/2πR. Thus

B₁ = μ₀I₁/2πR₁= (4 × 10^-7 T-m/A)(30 A)/{2π(0.25 m)} = 2.40 × 10^-5 T,

B₂ = μ₀I₂/2π{r₁+r₃} = (4 × 10^-7 T-m/A)(22 A)/{2π(0.25 m + 0.70 m)} = 4.632 × 10^-6 T,

The net magnetic field is thus

B_net = B₂ – B₁ = -1.937 × 10^-5 T . The minus sign indicates that it is directed down.

Point B:

The magnitude of the magnetic field due to a wire is given by B = μ₀I/2πR. Thus

B₁ = μ₀I₁/2πR₂ = (4 × 10^-7 T-m/A)(30 A)/{2π(0.35 m)} = 1.714 × 10^-5 T,

B₂ = μ₀I₂/2π{r₃-r₂} = (4 × 10^-7 T-m/A)(22 A)/{2π(0.70 m – 0.35 m)} = 1.257 × 10^-5 T,

The net magnetic field is thus

B_net = B₁ + B₂ = +2.971 × 10^-5 T . The plus sign indicates that it is directed up.

Point C:

The magnitude of the magnetic field due to a wire is given by B = μ₀I/2πR. Thus

B₁ = μ₀I₁/2π{r₃+r₄} = (4 × 10^-7 T-m/A)(30 A)/{2π(0.70 m + 0.50 m)} = 5.000 × 10^-6 T,

B₂ = μ₀I₂/2πR₄ = (4 × 10^-7 T-m/A)(22 A)/{2π(0.50 m)} = 8.800 × 10^-6 T,

The net magnetic field is thus

B_net = B₁ – B₂ = -3.80 × 10^-6 T . The minus sign indicates that it is directed down.

Determine the direction and magnitude of the net magnetic field at point A and B due to the two wires shown below. Wire 1 carries a current I₁ = 650 mA and I₂ = 475 mA. Point A is a distance r₁ = 1.20 m from wire 1 and point B is r₂ = 2.20 m away.

Point A:

The magnitude of the magnetic field due to a wire is given by B = μ₀I/2πR. Thus

B₁ = μ₀I₁/2πR₁ = (4 × 10^-7 T-m/A)(0.65 A)/{2π(1.20 m)} = 1.0833 × 10^-7 T ,

B₂ = μ₀I₂/2πR₂ = (4 × 10^-7 T-m/A)(0.475 A)/{2π(2.20 m)} = 4.3182 × 10^-8 T .

Thus B_net = -i B₁ + j B₂ = -i[1.0833 × 10^-7] + j[4.3182 × 10^-8].

Using the Pythagorean Equation,

B_net = [(B₁)²+(B₂)²]^½ = [(1.0833× 10^-7)²+(4.3182× 10^-8)²]^½ = 1.1662 × 10^-7 T .

The angle is given by

θ = arctan(|B₂/B₁|) = arctan(|0.43182/1.0833|) = 21.7° .

The net magnetic field at point A is 1.17 × 10^-7 T at 158.3° above the line formed by r₂.

Point B:

The magnitude of the magnetic field due to a wire is given by B = μ₀I/2πR. Thus

B₁ = μ₀I₁/2πR₂ = (4 × 10^-7 T-m/A)(0.65 A)/{2π(2.20 m)} = 5.9091 × 10^-8 T ,

B₂ = μ₀I₂/2πR₁ = (4 × 10^-7 T-m/A)(0.475 A)/{2π(1.20 m)} = 7.9166 × 10^-8 T .

Thus B_net = -i B₂ + j B₁ = -i[7.9166 × 10^-8] + j[5.9091 × 10^-8] .

Using the Pythagorean Equation,

B_net = [(B₁)²+(B₂)²]^½ = [(5.9091× 10^-8)²+(7.9166× 10^-8)²]^½ = 9.879 × 10^-8 T .

The angle is given by

θ = arctan(|B₂/B₁|) = arctan(|5.9091/7.9166|) = 36.7° .

The net magnetic field at point A is 9.88 × 10^-8 T at 143.3° above the line formed by r₂.

In the diagram below, the cross-section of two wires is shown. The wire on the left carries a current of I₁ = 12.5 A directed into the paper while the left has current I₂ = 8.5 A directed out of the paper. Both wires are R = 0.45 m from point A and the θ = 90°. What is the direction and magnitude of the magnetic field at A?

The field due to a wire is perpendicular to a radial vector from the wire. The direction is given by the right hand rule.

The magnitude of the magnetic field is

B₁ = μ₀I₁/2πR = (2 × 10^–7 Tm/A)(12.5 A)/(0.45 m) = 5.556 × 10^–6 T .

and

B₂ = μ₀I₂/2πR = (2 × 10^–7 Tm/A)(8.5 A)/(0.45 m) = 3.778 × 10^–6 T .

Since θ = 90° and we are given an isosceles triangles, then α = 45°, and thus B₁ and B₂ are 45° below horizontal as shown. So

B₁ = –i (5.556 × 10^–6 T)cos45° – j (5.556 × 10^–6 T)sin45° = –i (3.929 × 10^–6 T) – j(3.929 × 10^–6 T)

and

B₂ = i (3.778 × 10^–6 T)cos45° – j (3.778 × 10^–6 T)sin45° = i (2.671 × 10^–6 T) – j(2.671 × 10^–6 T)

Thus we find the net magnetic field to be

B_net = B₁ + B₂ = –i (1.258 × 10^–6 T) – j(6.600 × 10^–6 T)

A +6.00 πC charge is moving with a speed of 7.50 × 10⁶ m/s parallel to a long, straight wire. The wire carries a current of 67.0 A in a direction opposite to that of the moving charge, and is 5.00 cm from the charge. Find the magnitude and direction of the force on the charge.

We sketch a lengthwise and a side view of the problem.

Viewed from above, the wire creates a magnetic field which is into the paper on the right hand side but out on the left hand side. Using the Right Hand Rule, we find that the force on the positive charge is out away from the wire. The magnitude of the field created by the wire at the location of the charge is

B = μ₀I/2πr = (4 × 10^-7 T-m/A)(67 A)/{2π(0.05 m)} = 2.68 × 10^-4 T .

Thus the force on the charge is given by

F = qvBsin(θ) = (6 × 10^-6 C)(7.50 × 10⁶ m/s)(2.68 × 10^-4 T)sin(90°) = 1.21 × 10^-2 N .

The charge experiences a force of 1.21 × 10^-2 N directed away from the wire.

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 0.85 m, while the mass of each is 0.073 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.2 × 10^-3 m. Determine the current in the rods.

We sketch a lengthwise and a end-on view of the problem.

The bottom rod floats because a magnetic force, F_m, balances its weight. The magnetic force exists because the bottom rod carries a current in a magnetic field, B, generated by the upper current-bearing rod. Applying Newton's Second Law, we find

F_m - mg = 0 .	(1)

The magnetic force is The magnitude of the magnetic field is

F_m = ILBsin(90°) .	(2)

The strength of the magnetic field is given by

B = μ₀I/2πr .	(3)

Substituting equation (3) into equation (2) and the result into equation (1), we find

IL[μ₀I/2πr]sin(90°) - mg = 0 .

Solving for I, this reduces to

I = [mg2πr/μ₀L]^½ .

With the given values, this becomes

I = [(0.793 kg)(9.81 m/s²)2π(8.2 × 10^-3 m)/(4π × 10^-7 T-m/A)(0.85 m)]^½ = 186 A .

The current in each rod is 186 Amperes.

An electron is travelling 10.0 cm above and parallel to a long thin wire carrying 8.00 A of current. The conventional current is out of the paper as shown and the wire is parallel to the surface of the earth. What is the wire's magnetic field at the electron location? With what speed and direction must the electron be moving such that the magnetic force exactly balances the force due to gravity? The charge of an electron is -1.60 × 10^-19 C and its mass is 9.11 × 10^-31 kg.

In the end view, we can use the right hand rule to determine that the magnetic field due to the wire is to the left. Since the weight acts down, we want the magnetic force to act up as shown. Since F_m and B are perpendicular to v, the electron can only be moving parallel or antiparallel to the current. Using the right hand rule, a positive charge moving into the page feels an upward force, thus the electron must be moving out of the page.

We want F_m = W, or

qvB = mg .	(1)

We know that the field due to the wire is given by

B = μ₀I/2πR .	(2)

Combining (1) and (2) and isolating v we get

v	= 2πmgR / μ₀Iq
	= (2π) (9.11 × 10^-31 kg)(9.81 m/s²)(0.10 m) / (4π × 10^-7 Tm/A)(8 A)(1.602 × 10^-19 C)
	= 3.5 × 10^-6 m/s

Two current carrying wires are parallel to each as shown in diagram (i) below. The side view is given in diagram (ii). The wires are 0.35 m apart. The current in the first wire is 25 A and 18 A in the second. The wires are 15.0 m long. What is the magnetic field (magnitude and direction) at the second wire due to the first wire? What is the force (magnitude and direction) on the second wire because of the magnetic field?

We are asked to consider the effects of wire #1 on wire #2. Using the Right Hand Rule, Wire #1 creates a magnetic field directed into the paper at the location of wire #2. The magnitude of this field is

B₁ = μ₀I₁/2πr = (4π × 10^-7 T-m/A)(25 A)/{2π(0.35 m)} = 1.429 × 10^-5 T .

Using the Right Hand Rule, the second wire feels a magnetic force directed toward the first wire. The magnitude of this force is given by

F₂₁ = I₂LB₁sin(q) = (18 A)(15.0 m)(1.429 × 10^-5 T)sin(90°) = 3.857 × 10^-3 N .

So the second wire experiences a force of 3.86 × 10^-3 N towards the first wire because both carry currents.

We could also consider the effects of wire #2 on wire #1. Using the Right Hand Rule, Wire #1 creates a magnetic field directed out the paper at the location of wire #1. The magnitude of this field is

B₂ = μ₀I₂/2πr = (4π × 10^-7 T-m/A)(18 A)/{2π(0.35 m)} = 1.029 × 10^-5 T .

Using the Right Hand Rule, the first wire feels a magnetic force directed toward the second wire. The magnitude of this force is given by

F₁₂ = I₁LB₂sin(θ) = (25 A)(15.0 m)(1.029 × 10^-5 T)sin(90°) = 3.857 × 10^-3 N .

Not unexpectedly, each wire experiences an equal but opposite force.

A wire carrying a current I is shaped as shown below. Find the magnitude of the magnetic field at point P using the Law of Biot-Savart. The identity ∫dx [x²+b²]^-3/2 = x/(b²[x²+b²]^½ + C) may be of assistance.

We label the pieces of the wire A, B, and C. The field due to a segment of infinitesimal piece of wire is given by

dB = (μ₀/4π)I [dl × (r/r³)] ,

where r is the vector from the segment to the point of interest. For the radial pieces A and C, dl = dr and thus dl × (r/r) = 0. We need only consider the contribution from B.

Next we choose a set of axis and pick an arbitrary piece of segment B.

We see that dl = idx, r = -xi - hj, and r = [x²+h²]^½. The cross product yields

dl ´ r = idx ´ (-xi- hj) = -khdx .

The expression for the magnetic field is thus,

dB = -k (μ₀Ih/4π)dx/[x²+h²]^3/2 .

We integrate over the entire length of the segment to get B

B	= ∫_-a^L-a dB
	= -k (μ₀I/4π) ∫_-a^L-a h/[h² + x²]^3/2 dx
	= -k (μ₀I/4π){x / h[h² + x²]^½ \|_-a^L-a}
	= -k (μ₀I/4πh){(L-a)/ [h² + (L-a)²]^½ + a/[h² + a²]^½}

Note, in the limit L >> h and a >> h, that this reduces to the familiar result for a field due to a long wire

B_wire = (μ₀/4π)(I/h) .

A wire carrying a current I is shaped as shown below. The arcs are circular of radii a and b. The straight pieces a radial to the centre of the shape. Find the magnitude of the magnetic field at the centre of the shape using the Law of Biot-Savart. The relationship S = rq may be of use.

The field due to a segment of infinitesimal piece of wire is given by

dB = (π₀/4π)I [dl × (r/r³)] ,

where r is the vector from the segment to the point of interest. For the radial pieces in the diagram above, dl = dr and thus dl × r = 0. We need only consider the contribution from curved arcs.

The field due to each side curve is the same by symmetry and is directed into the paper at the origin using the right hand rule. The field due to the top and bottom curves is the same by symmetry and is also directed into the paper at the origin using the right hand rule.

Consider a current carrying arc,

Since dl is perpendicular to r and r = R, dl ´ (r/r³) = -kdθ/R. Thus the magnetic field is

dB = -k (μ₀/4π)(I/R)dθ .

We integrate over the entire angular extent of the segment to get B

B	= ∫_-α^-α dB
	= -k (μ₀/4π)(I/R) ∫_-α^-αdθ
	= -k (μ₀/4π)(I/R) { \|_-α^-α}
	= -k (μ₀I/2π)(α/R)

For the top and bottom coils, R = b, and α = 2π/3. For the side coils R = a and α = π/3. Thus the total field is

B = -k 2(μ₀I/2π) {(2π/3b) + (π/3a)} = -k (μ₀I/3)(2/b + 1/a) .

Note that in the limit a = b, this reduces to the familiar result for a field due to a circular wire

B = μ₀I/a .

A loop of wire has the shape of two concentric semicircles connected by two radial segments. The loop carries current I as shown. Find the magnetic field at the point P using the Law of Biot-Savart.

The field due to a segment of infinitesimal piece of wire is given by

dB = (π₀/4π)I [dl × (r/r³)] ,

where r is the vector from the segment to the point of interest. For the radial pieces in the diagram above, dl = dr and thus dl × r = 0. We need only consider the contribution from curved arcs.

The field due to the large arc out of the paper at the point P using the right hand rule. The field due to the small arc is directed into the paper at the P. From question 3, we found that the field due to an arc was

B = (μ₀I/2π)(α/r) .

For both arcs, α = π/2. For the top arc r = 2R, and for the bottom r = R. Taking out of the page as positive

B = (μ₀I/4)(1/2R – 1/R) = –μ₀I/8R .

The net field is into the paper.

A long copper wire of cross-sectional radius R carries a uniform current I. Use Ampere's Law to determine B as a function of the distance a from the centre of the wire. Sketch the result.

For cases of radial symmetry, Ampere's Law

∫_C B • dl = μ₀I_enclosed ,

reduces to

2πrB = 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).

(i) r < R

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

I_enclosed = I(πr²/πR²) = I (r²/R²) .

Thus (1) becomes

2πrB = μ₀ I (r²/R²) ,

and we find

B = μ₀Ir / 2πR² .

(ii) r > R

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

Thus (1) becomes

2πrB = μ₀ I,

and we find

B = μ₀I / 2πr .

which is the start result outside a wire.

A graph of B as a function of r looks like

Note the kink (change in slope) in the graph at r = R where the solid ends.

A coaxial cable consists of two concentric very thin, very long, cylindrical shells of radius R and 2R. The shells carry equal and opposite currents I. Determine the magnetic field B as a function of the distance a from the centre of the wire. Sketch the result.

For cases of radial symmetry, Ampere’s Law

∫_C B•dl = μ₀I_enclosed ,

reduces to

2π aB = 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 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).

(i) a < R

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

(ii) R < a < 2R

Here the enclosed current is I, so B = μ_oI/2πa.

(iii) a > 2R

Here we have no enclosed current, I_enclosed = +I + –I = 0, so B is exactly zero.

A graph of B as a function of a looks like

Note the jumps at shells of current.

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. Use Ampere's Law to find the magnetic fields as a function of radial distance from the centre of the pipe. Sketch the result.

Note that the area of the pipe is A = π(2R)² - π(R)² = 3πR².

For cases of radial symmetry, Ampere's Law

∫_C B • dl = μ₀I_enclosed ,

reduces to

2πrB = 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).

(i) r < R

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

(ii) R < r < 2R

The amount of current passing through the amperian circle is proportional to the area of the wire it encloses. The area of wire inside the circle is A = πr² - πR². Thus

I_enclosed = IA_enclosed / A_total = I (πr² – πR²)/(3πR²) = I (r² – R²)/(3R²) .

Thus (1) becomes

2πrB = μ₀ I (r² – R²)/(3R²) ,

and we find

B = (μ₀I / 2π)(r² – R²)/(3rR²) .

(iii) r > 2R

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

Thus (1) becomes

2πrB = μ₀ I,

and we find

B = μ₀I / 2πr .

which is the start result outside a wire.

A graph of B as a function of r looks like

Note the kinks (changes in slope) in the graph at r = R and r = 2R where the solid begins and ends.

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. We have a thick shell in this problem so note that the area of the shell is A_shell = [π(r₃)² - π(r₂)²].

For cases of radial symmetry, Ampere's Law

∫_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.

(i) 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 proportional to the area

I_enclosed = I(πa²/πr₁²) = I (a²/r₁²) .

Thus (1) becomes

2πaB = μ₀I (a²/r₁²),

and we find

B = μ₀Ia / 2πr₁² .

(ii) 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 .

(iii) 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

I_enclosed = I – j_shell(πa² – πr₂²) = I – I[(a² – r₂²)/(r₃² – r₂²)] .

Thus (1) becomes

2πaB = μ₀ I[1 – (a² – r₂²)/(r₃² – r₂²)] ,

and we find

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

(iv) 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 a looks like

Note the kinks (changes in slope) in the graph at r₁, r₂, and r₃ which are the boundaries of the solids.

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 ∫ e^-axxdx = -(x/a)e^-ax + e^-ax/a + C may be of use.

For cases of radial symmetry and nonconstant j(r), Ampere's Law

∫_C B • dl = μ₀I_enclosed ,

reduces to

2πaB = 2πμ₀ ∫₀^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.

(i) 0 < a < R

Equation (1) reduces to

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

(ii) a > R

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

B	= (μ₀/a) ∫₀^R j(r)rdr
	= (μ₀A/a) ∫₀^R e^-Krrdr
	= (μ₀A/a){–(r/K)e^-Kr + (1/K) e^-Kr \|₀^R }
	= (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 = 2A(1 – e^-KR – KRe^-KR)/K² .

A graph of B as a function of a looks like

Note the kink (change in slope) in the graph at the boundary of the solid.

Questions? mike.coombes@kwantlen.ca