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1. 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 òdx[x² + b²]^–3/2 = x/b²[x²+b²]^1/2 + 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)] ,          (1)

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 then must express dl and r in Cartesian coordinates. Here we have dl = i dx and r = –i x – j h. As well, r = | r | = [x² + h²]^½.



Thus equation (1) becomes

dB = (μ₀/4π)I [i ´ (–i x – j h)]dx/[x² + h²]^3/2 .

Doing the cross product yields

dB = –k (μ₀/4π)Ihdx/[x² + h²]^3/2 .

This tell us that the magnetic field at point P is directed into the page which agrees with what we get from the Right–Hand Rule.

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

B	= ò–aL–a dB
	= –k (μ0/4π)I ò–aL–a h/[h2 + x2]3/2 dx
	= –k (μ0/4π)I {x / h[h2 + x2]½ |–aL–a}
	= –k (μ0/4π)(I/h){ (L–a)/ [h2 + (L–a)2]½ + a/[h2 + a2]½}

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

B_wire = (μ₀/4π)(I/h) .



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2. A wire carrying a current I is shaped as shown below. The arcs are circular of radii a and b where a > b. The straight pieces are radial from the centre of the shape. Each large arc subtends an angle of 2π/3. Find the magnitude of the magnetic field at the centre of the shape using the Law of Biot–Savart. The relationship S = rθ may be of use.



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

dB = (μ₀/4π)I [dl ´ (r/r³)] ,          (1)

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.



In general we have arc segments to deal with. Symmetry suggests that we deal with the integrals involving circles and arcs in polar coordinates. Consider a current carrying arc such as is shown below where we have considered an arbitrary portion of the arc. We will end up integrating over θ from –α to +α . First however, we must express dl and r in Cartesian coordinates. This is a little trickier to do here than in the previous question so let’s move dl and r to the origin and do a little geometry and trigonometry. Remember vectors can be moved around as long as we do not change length or orientation.



Recall that the magnitude of dl is dl = | dl | = Rdθ . Thus dl = Rdθ [ i cos(θ) – j sin(θ) ]. As well, r = R [–i sin(θ) – j cos(θ)] since r = | r | = R. Thus equation (1) becomes

dB = (μ₀/4π)(I/R³){ Rdθ [ i cos(θ) – j sin(θ) ] ´ R [–i sin(θ) – j cos(θ)] } .

Working out the cross product yields

dB = –k (μ₀/4π)(I/R)dθ [cos²(θ) + sin²(θ)] = –k (μ₀/4π)(I/R)dθ .

So the magnetic field at the origin of the arc is directed into the page which is what we already found from the Right–Hand Rule.

Next, we integrate over the entire angular extent of the segment to get B

B	= ò–αα dB
	= –k (μ0/4π)(I/R) ò–α αdθ
	= –k (μ0/4π)(I/R) {θ |–α α}
	= –k (μ0/2π)(Iα/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/π a .



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3. 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 = (μ₀/4π)I [dl ´ (r/r³)] ,          (1)

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.



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4. An arc of thin wire of radius R lies in the positive quadrant of the xy plane. It carries a current I clockwise. Determine the integral expression for the magnetic field acting at x = a. Don't bother evaluating the integral.



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

dB = (μ₀/4π)I [dl ´ r] / r³ ,          (1)

where r is the vector from the segment to the point of interest. Putting the origin at the centre of the radius of curvature and picking an arbitrary piece or the wire dl, we find S = i Rcos(φ) + j Rsin(φ) and P = i a. Therefore r = P – S = i [a – Rcos(φ)] – j Rsin(φ). The magnitude of this vector is r = [a² – 2aRcos(φ) + R²]^½. Note that dl = –φ Rdφ , since the current moves in the clockwise or negative direction. To do the cross product, we need to express dl in ijk notation. Placing dl at the origin as shown in the next diagram, doing some geometry, and determining the components of dl, we see that dl = Rdφ [ i sin(φ) – j cos(φ)].



Thus equation (1) for this problem is

dB = (μ₀/4π)I Rdφ [i sin(φ)–j cos(φ)]´ [i [a–Rcos(φ)]–j Rsin(φ)]/[a²–2aRcos(φ)+R²]^3/2 .

Doing the cross product yields



And we have

dB = k (μ₀/4π)IRdφ [acos(φ) – R]/[ a²–2aRcos(φ)+R²]^3/2 .

This tells us that the magnetic field at point P is directed out of the page which agrees with what we get from the Right–Hand Rule.

We integrate over the entire length of the segment from 0 to π/2 to get B

B	= ò0½π dB
	= –k (μ0IR/4π) ò0½π dφ [acos(φ) – R]/[ a2–2aRcos(φ)+R2]3/2



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5. A flat ribbon of length L and width W lies in the yz plane as shown below. It carries surface current K = kCz. Determine the magnetic field P = (b, 0, 0).



The field due to a surface current is given by

B = (μ₀/4π) ò_surface dS [K ´ r] / r³ ,          (1)

where r is the vector from the segment of the surface current to the point of interest. Using the given origin and picking an arbitrary piece of the wire dS, we find S = j y + k z and P = i b. Therefore r = P – S = i b – j y – k z. The magnitude of this vector is r = [b² + y² + z²]^½. The surface element is dS = dydz where -½W £ y £ ½W and 0 £ z £ L. Thus equation (1) for this problem is

B = (μ₀/4π) ò_½W^½W dy ò₀^L dz {kCz ´ [i b – j y – k z]} / [b² + y² + z²]^3/2

Doing the cross product yields



And we have

B = (μ₀/4π)C ò_½W^½W dy ò₀^L dz {i zy + j zb} / [b² + y² + z²]^3/2 .

Using MAPLE to integrate the components of the magnetic field, we find B_x = 0 and





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6. A semi–circular ribbon has inner radius R and outer radius R + a. The ribbon carries a surface current K = Csφ where s is radial distance and φ is the angular unit vector in polar coordinates. Determine the magnetic field acting at the centre of the curvature.



The field due to a surface current is given by

B = (μ₀/4π) ò_surface dS [K ´ r] / r³ ,          (1)

where r is the vector from the segment of the surface current to the point of interest. The object has planar circular (polar coordinate) symmetry, so the variables of integration should be s and φ with the area element being dS = sdsdφ where R £ s £ R+a and 0 £ φ £ π . All directions should be given in standard ijk notation for convenience in dealing with the cross product. We know that φ = –i sin(φ) + j cos(φ). The origin is at the centre of curvature. Picking an arbitrary piece of the object dS, we find S = i x + j y and P = 0. Therefore r = P – S = – i x – j y. The magnitude of this vector is r = [x² + y²]^½. We need to express x and y in terms of s and φ . We know x = scos(φ) and y = ssin(φ), so r = – i scos(φ) – j ssin(φ) and r = s. Thus equation (1) for this problem is

B = (μ₀/4π) ò_R^R+a ds ò₀^π dφ s {Cs[–i sin(φ) + j cos(φ)] ´ [–i scos(φ) – j ssin(φ)]} / s³

Doing the cross product yields



And we have

B = k (μ₀/4π) C ò_R^R+a ds ò₀^π dφ = k (μ₀/4) CR .



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7. A flat disk of radius R lies in the xy plane as shown below. It carries surface charge density Σ . The disk is given angular velocity Ω φ where φ is the cylindrical rotational unit vector. Since the charge is moving, we now have a surface current density K = Σ sΩ φ where s is the distance from the centre of the disk. Determine the magnetic field acting at P = (0, 0, b).



The field due to a surface current is given by

B = (μ₀/4π) ò_surface dS [K ´ r] / r³ ,          (1)

where r is the vector from the segment of the surface current to the point of interest. The object has planar circular (polar coordinate) symmetry, so the variables of integration should be s and φ with the area element being dS = sdsdφ where 0 £ s £ R and 0 £ φ £ 2π . All directions should be given in standard ijk notation for convenience in dealing with the cross product. We know that φ = –i sin(φ) + j cos(φ). The origin is at the centre of the disk. Picking an arbitrary piece of the disk dS, we find S = i x + j y and P = k b. Therefore r = P – S = – i x – j y + k b. The magnitude of this vector is r = [x² + y² + b²]^½. We need to express x and y in terms of s and φ . We know x = scos(φ) and y = ssin(φ), so r = – i scos(φ) – j ssin(φ) + k b and r = [s² + b²]^½. Thus equation (1) for this problem is

B = (μ₀/4π)ò₀^R ds ò₀^2πdφ s{Σ sΩ[–i sin(φ)+jcos(φ)]´ [–iscos(φ)–jssin(φ)+kb]}/[s²+b²]^3/2

Doing the cross product yields



And we have

B = (μ₀/4π) Σ Ω ò₀^R ds ò₀^2π dφ s² [i bcos(φ) + j bsin(φ) + k s] / [s² + b²]^3/2

The i and j terms vanish upon integration over φ , leaving





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8. A sphere of radius R is shown below. It carries volume charge density ρ . The sphere is given angular velocity Ω φ about the z axis where φ is the spherical rotational unit vector. Since the charge is moving, we now have a volume current density J = ρ rsin(θ)Ω φ where r is the radial distance from the centre of the sphere and angle θ is measured from the z–axis. Determine the magnetic field acting at P = (0, 0, b).



The field due to a volume current is given by

B = (μ₀/4π) ò_volume dV [J ´ g] / g³ ,          (1)

where g is the vector from the segment of the surface current to the point of interest. The object has spherical symmetry, so the variables of integration should be r, θ , and φ with the volume element being dV = r²sin(θ)drdθ dφ where 0 £ r £ R, 0 £ θ £ π and 0 £ φ £ 2π . All directions should be given in standard ijk notation for convenience in dealing with the cross product. We know that φ = –i sin(φ) + j cos(φ). The origin is at the centre of the sphere. Picking an arbitrary piece of the sphere dV, we find S = i x + j y + k z and P = k b. Therefore g = P – S = – i x – j y + k (b – z). The magnitude of this vector is g = [x² + y² + z² + b² – 2bz]^½. We need to express x, y and z in terms of r, θ , and φ . We know x = rsin(θ)cos(φ), y = rsin(θ)sin(φ), and z = rcos(θ) so r = – i rsin(θ)cos(φ) – j rsin(θ)sin(φ) + k [b – rcos(θ)] and g = [r² + b² – 2br cos(θ)]^½. Thus equation (1) for this problem is

B = (μ₀/4π) ò₀^R dr ò₀^2π dφ ò₀^π dθ r²sin(θ) {ρ rsin(θ)Ω [–i sin(φ)+jcos(φ)]´ [– i rsin(θ)cos(φ) – j rsin(θ)sin(φ) + k (b – rcos(θ))]}/[r² + b² – 2br cos(θ)]^3/2

Doing the cross product yields



And we have



The i and j terms vanish upon integration over φ , leaving

B_z = (μ₀/2) ρ Ω ò₀^R dr ò₀^π dθ r⁴ sin³(θ) / [r² + b² – 2brcos(θ)]^3/2 .

The integrand may be converted to a series by expanding about s = 0. Only the first term gives a non-zero result for the integration over θ . We are left with

B_z = 2μ₀ ρ ΩR⁵ / 15b³ .



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9. A cylinder has its central axis oriented in the z direction. Its base is in the xy plane. The base has radius R and infinite height. It carries a current density J = Csz, where s is the cylindrical radial distance. Determine the magnetic field acting at a distance b from the centre of the base.



The field due to a volume current is given by

B = (μ₀/4π) ò_volume dV [J ´ r] / r³ ,          (1)

where r is the vector from the segment of the surface current to the point of interest. The object has cylindrical symmetry, so the variables of integration should be s, φ , and z. All directions should be given in standard ijk notation for convenience in dealing with the cross product. We take the origin is at the bottom of the cylinder and assume that the point of interest is on the y axis. Picking an arbitrary piece of the cylinder dV = sdsdφ dz, we find S = i x + j y + k z and P = j b. Therefore r = P – S = – i x + j (b – y) – k z. The magnitude of this vector is r = [x² + y² + z² + b² – 2by]^½. We need to express x, y and z in terms of s, φ , and z. We know x = scos(φ), y = ssin(φ), and z = z so r = – i scos(φ) + j [b – ssin(φ)] – k z and r = [z² + s² + b² – 2bs sin(φ)]^½. Thus equation (1) for this problem is



The cross product yields



Thus our equation becomes



The j term vanishes leaving



The integral over z yields



The integrand may be converted to a series by expanding about s = 0. Only the first term gives a non-zero result for the integration over φ . We are left with

B_z = – μ₀CR³/6b .

Noting that the current in this case I = ò₀^R dr ò₀^2π dφ sJ = 2π CR³/3 , the above answer reduces to B_z = – μ₀I / 4π b. The same result as a long thin wire.



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10. Which of the following are possible magnetic fields?
1. B_x = 2xy, B_y = 2z+3, and B_z = 5–2yz.
2. B_x = x²+y², B_y = 3, and B_z = –xz.
3. B_x = y³–z³, B_y = z³–x³, and B_z = x³–y³ .
4. B_x = ln(x), B_y = y/x, and B_z = 2y/x.

Real magnetic fields have zero divergence, Ñ·B = 0

(a)	Ñ ·B	= ¶ Bx/¶ x + ¶ By/¶ y + ¶ Bz/¶ z
		= ¶ (2xy)/¶ x + ¶ (2z + 3)/¶ y + ¶ (5 – 2yz)/¶ z
		= 2y + 0 – 2y
		= 0

So this may be a magnetic field.

(b)	Ñ ·B	= ¶ Bx/¶ x + ¶ By/¶ y + ¶ Bz/¶ z
		= ¶ (x2 + y2)/¶ x + ¶ (3)/¶ y + ¶ (–xz)/¶ z
		= 2x + 0 – x
		= x

So this cannot be a magnetic field.

(c)	Ñ ·B	= ¶ Bx/¶ x + ¶ By/¶ y + ¶ Bz/¶ z
		= ¶ (y3 – z3)/¶ x + ¶ (z3 – x3)/¶ y + ¶ (x3 – y3)/¶ z
		= 0 + 0 + 0
		= 0

So this may be a magnetic field.

(d)	Ñ ·B	= ¶ Bx/¶ x + ¶ By/¶ y + ¶ Bz/¶ z
		= ¶ ln(x)/¶ x + ¶ (y/x)/¶ y + ¶ (–2y/x)/¶ z
		= 1/x + 1/x + 0
		= 2/x

So this cannot be a magnetic field.



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11. The diagrams below show a current–carrying loop of radius a located in the xz plane and a point P located in the yz plane. As shown in Figure 1, point P is a distance R from the centre of the loop at an angle α to the y axis. As shown in figure 2, point A is a point on the loop at an angle β from the z axis. Vector r = P – A is the distance from points A to P.

1. Obtain expressions for P, A, r, and dl in Cartesian (ijk) coordinates.
2. Find dl ´ r in Cartesian coordinates.
3. Use the Law of Biot–Savart to obtain integral expressions for the Cartesian components of the magnetic field at point P.
4. Assume R >> a and use the identity (1 – x)^z = 1 + zx for small x to expand and simplify the integrands in part (c). Integrate to show

   

   where m = Iπ a² is the dipole moment of the loop.

5. Figure 3 shows the relationship between the k and j axes and the r and θ axes. Use this relationship to show that







1. Expressing the vectors in Cartesian form.

Examining Figure 2, we immediately see that

A = asin(β ) i + 0 j + acos(β ) k .

By geometry, we see that dl is angle β below the positive x axis. As well, the length of the arc is dl = | dl | = a dβ . Hence

dl = adβ cos(β ) i + 0 j - adβ sin(β ) k .

We are told P is in the yz plane so

P = 0 i + Rcos(α) j + Rsin(α) k .

Vector r is the vector from A to P, so

r = P - A = -asin(β ) i + Rcos(α) j + [Rsin(α) - acos(β )] k .

2. The cross product dl ´ r is

dl ´ r	= adβ {cos(β ) i + 0 j - sin(β ) k} ´ {-asin(β ) i + Rcos(α) j + [Rsin(α)-acos(β )] k}
	= adβ { Rcos(α)sin(β ) i + [a- Rsin(α)cos(β )] j + Rcos(α)cos(β ) k }

3. The Law of Biot-Savart states that

dB = (μ₀/4π)(I/r³) dl ´ r .

We first need to find r = | r |,



Then we observe that we need to integrate around the entire circumference of the loop from β = 0 to β = 2π . So the field is



4. These are messy integrals. However, an examination of the symmetry if the problem leads up to expect that B_x must be zero as contributions to from opposite sides of the loop will cancel. Using MAPLE confirms that B_x = 0.

Expanding the integral’s denominator, we have



We neglect terms of order (a/R)² and higher. Thus the x and y components of the magnetic field are



and



where we have used MAPLE to evaluate the integrals.

5. Examining Figure 3, we see that the magnetic field lying along the r axis is

B_r = B_ycos(α) + B_zsin(α) .

After using some trigonometric identities, we find

B_r = 2μ₀mcos(α)/4πR² .

Similarly, the magnetic field along the θ axis is

B_θ = -B_ysin(α) + B_zcos(α) ,

which reduces to

B_θ = μ₀msin(α)/4π R² .

The B_φ term is zero since it is at right angles to B_r and B_θ which in turn makes it perpendicular to the B_y and B_z terms.

Questions?mike.coombes@kwantlen.ca