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1. The earth's magnetic field is approximately 5 ´ 10^–5 T. If an electron is traveling perpendicular to the field at 1000 m/s, determine its cyclotron radius and frequency. Ignore the effects of gravity and the electric field of the atmosphere. The mass of an electron is 9.11 ´ 10^–31 kg.

First we sketch the behaviour or the electron, assuming that the magnetic field points out of the paper.



The magnetic force is F = qv ´ B. Since the velocity and magnetic field are at right angles, F = evB. The force is directed to the centre of the circle, so it is the centripetal force. Applying Newton’s Second Law

evB = mv²/R .           (1)

Solving (1) for R, we get

R = mv/eB = (9.11´ 10^-31 kg)(1000 m/s)/(1.602´ 10^-19 C)(5 ´ 10^-5 T) = 1.1 ´ 10^-4 m/s .

For an object to travel in a circle with constant speed

v = 2π R/T = 2π Rf ,

where T is the period and f is the cyclotron frequency. Thus

f = v/(2π R) .           (2)

From (1), v = eBR/m, thus

f = eB/2π m = (1.602´ 10^-19 C)(5´ 10^-5 T)/(2π )(9.11´ 10^-31 kg) = 1.4 MHz.

Note that the cyclotron frequency is independent of the velocity of the charge.



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2. A beam of protons moves in a circle of radius 0.25 m. The beam moves perpendicular to a 0.30 T magnetic field. (a) What is the speed of each proton? (b) Determine the cyclotron frequency of the protons. (c) Determine the magnitude of the centripetal force on each proton. Protons have mass 1.673 ´ 10^–27 kg.

First we sketch the behaviour or the protons, assuming that the magnetic field points out of the paper.



The magnetic force is F = qv ´ B. Since the velocity and magnetic field are at right angles, F = evB. The force is directed to the centre of the circle, so it is the centripetal force. Applying Newton’s Second Law

qvB = mv²/R .

Solving for v, we get

v = qBR/m = (1.602´ 10^-19 C)(0.30 T)(0.25 m)/(1.673´ 10^-27 kg) = 7.226 ´ 10⁶ m/s .

Note that the magnitude of the charge of a proton is the same as that of an electron.

For an object to travel in a circle with constant speed

v = 2π R/T = 2π Rf ,

where T is the period and f is the cyclotron frequency. Thus

f = v/(2π R) = qB/2π m = (1.602´ 10^-19 C)(0.30 T)/(2π )(1.673´ 10^-27 kg) = 4.6 MHz.

As well,

F =qvB = (1.609´ 10^-19 C)( 7.226´ 10⁶ m/s)(0.30 T) = 3.488 ´ 10^-13 N .



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3. A velocity selector with an electric field of 4.50 ´ 10³ V/m and a magnetic field of 0.100 T is used to select the speed of an ion of charge +e before it enters a mass spectrometer. A 0.400–T magnetic field then bends the ion into a circular path of radius 0.230 m. What is the mass of the ion? What is the atomic mass of the ion? What element is the ion? (Atomic mass is the mass in grams of one mole of the substance. N_A = 6.022 ´ 10²³)

In a velocity selector, the electric force exactly balances the magnetic force, qE = qvB₁, yielding an equation for the velocity of the charge

v = E/B₁ ,           (1)

where B₁ is the magnetic field strength in the selector.

In the mass spectrometer, the charge is bent into a circular path. The radius of the path, as derived in question 6, is

R = mv/qB₂ ,           (2)

where B₂ is the magnetic field strength in the spectrometer. Combining (1) and (2) yields an equation for the mass of the charge

m = qB₁B₂R/E .

A singly ionized atom has the same magnitude of charge as the electron it lost. Using the given information,

m = (1.602´ 10^-19 C)(0.1 T)(0.4 T)(0.230 m)/(4500 V/m) = 3.275 ´ 10^-25 kg .

The ion has a mass of 3.28 ´ 10^-25 kg.

The mass of one mole of these ions is

M = N_Avagadro m = (6.022 ´ 10²³)( 3.275 ´ 10^-25 kg) = 0.197 kg = 197 g .

Examining a Periodic Table, one finds that Au, gold, has this atomic mass.



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4. Suppose that an ion source produces doubly ionized gold ions (Au⁺⁺), each with a mass of 3.27 ´ 10^–25 kg. The ions are passed through a velocity selector with E = 5.00 ´ 10³ V/m and B = 0.150 T. Then, a 0.500–T magnetic field causes the ions to follow a circular path. Determine the radius of the path.

From question 4, we have

v = E/B₁ ,           (1)

where B₁ is the magnetic field strength in the selector, and

R = mv/qB₂ ,           (2)

where B₂ is the magnetic field strength in the spectrometer. Combining (1) and (2) yields an equation for the radius

R = mE/qB₁B₂ .

A doubly-ionized atom has the same magnitude of charge as the electrons it lost, i.e. 2e. Using the given information,

R = (3.27´ 10^-25 kg)(5000 V/m) / (2 ´ 1.602´ 10^-19 C)(0.15 T)(0.5T) = 0.068 m .

The radius of the path is 6.8 cm.



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5. Consider a thin wire carrying a constant current I. The wire has length L and lies in the xy plane at an angle of 45° to the x axis. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the thin wire.



The force on the wire is given by F = I ò_L dl ´ B .

Here, the infinitesimal piece of the wire dl = i dx + j dy and B = i 2xy + j 3 + k 5. The cross-product is



The x component of the force is thus

F_x = 5I ò₀^{L/Ö 2} dy = 5IL/Ö 2 .

The y component is

F_y = –5I ò₀^{L/Ö 2} dx = –5IL/Ö 2 .

And the z component yields

Fz	= I ò[3dx – 2xydy]
	= 3I ò0L/Ö 2 dx – 2I ò0L/Ö2[y]y dy
	= 3IL / Ö 2 – 2I y3/3 |0L/Ö 2
	= 3IL / Ö 2 – I L3 / 3Ö 2

Note that we used the fact that x = y when doing the y integral above.



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6. An arc of thin wire of radius R lies in the positive quadrant of the xy plane. It carries a current I clockwise. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the thin wire.

The force on the wire is given by F = I ò_arc dl ´ B . We need to spend some time expressing the infinitesimal piece of the wire dl in ijk notation.



Now dl has length Rdθ . Thus the vector is dl = i Rsin(θ)dθ – j Rcos(θ)dθ . As well, the magnetic field is B = i 2R²cos(θ)sin(θ)+ j 3 + k 5, where we have expressed x and y in terms of θ . Note that the arc runs from θ = π/2 to θ = 0. The cross-product is



The x component of the force is thus

F_x = –5IR òπ_/2⁰ cos(θ)dθ = –5IR sin(θ)|π_/2⁰ = 5IR .

The y component is

F_y = –5IR òπ_/2⁰ sin(θ)dθ = 5IR cos(θ)|π_/2⁰ = 5IR .

And the z component yields

Fz	= I òπ/20 [3Rsin(θ)+ 2R3 cos2(θ)sin(θ)] dθ
	= –3IR cos2(θ)|π/20 – (2/3)IR3 cos3(θ)|π/20
	= –3IR – (2/3)IR3



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7. A flat ribbon of length L and width W lies in the yz plane as shown below. It carries surface current K = kCz. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the ribbon.



The force on the ribbon is given by F = ò_ribbon K ´ B dS . In the yz plane, B = i 0 + j (2z+3) + k (5–yz). The cross product is



The x component of the force is thus

Fx	= –C ò–½W ½W dy ò0L dz [2z2 + 3z]
	= –C[ y|–½W ½W][(2/3)z3+(3/2)z2|0L ]
	= –CW[(2/3)L3+(3/2)L2] .



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8. A semi–circular ribbon has inner radius R and outer radius R + a. It lies in the xy plane centered on the origin. The ribbon carries a surface current K = Crθ , where θ is the polar coordinate unit vector. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the ribbon.



The force on the ribbon is given by F = ò_ribbon K ´ B dS . We need to express K in ijk notation, which involves determining the relationship between θ and i and j.



Now θ has unit length, thus we have θ = –i sin(θ)+ j cos(θ). As well, the magnetic field is B = i 2r²cos(θ)sin(θ)+ j 3 + k 5, where we have expressed x and y in terms of r and θ . The surface element dS = rdrdθ for polar coordinates. The cross-product is



The x component of the force is thus

F_x = 5C ò_R ^R+a dr ò₀^πdθ r² cos(θ)= 5C[ r³/3|_R ^R+a][ sin(θ)|₀^π] = 0 .

The y component is

F_y = 5C ò_R ^R+a dr ò₀^πdθ r² sin(θ)= 5C[ r³/3|_R ^R+a][–cos(θ)|₀^π] = 10C(R²a + Ra² + a³/3).

And the z component yields

Fz	= –C òR R+adr ò0πdθ [3r2 sin(θ)+ 2r4 cos2(θ)sin(θ)]
	= –3C[ r3/3|R R+a][–cos(θ)|0π] – 2C[ r5/5|R R+a][–cos3(θ)|0π]
	= –2C[(R+a)3–R3] – (4/5)C[(R+a)5–R5]



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9. 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. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the disk.



The force on the disk is given by F = ò_disk K ´ B dS . We need to express K in ijk notation, which involves determining the relationship between φ and i and j.



Now φ has unit length, thus we have φ = –i sin(φ) + j cos(φ). As well, the magnetic field is B = i 2s²cos(φ)sin(φ) + j 3 + k 5, where we have expressed x and y in terms of r and θ. The surface element dS = sdsdφ for cylindrical coordinates. The cross product is



The x component of the force is thus

F_x = 5Σ ò₀ ^R ds ò₀^2πdφ s² cos(φ) = 5Σ [ s³/3|₀ ^R][ sin(φ)|₀^2π] = 0 .

The y component is

F_y = 5Σ ò₀ ^R ds ò₀^2πdφ s² sin(φ) = 5C[ r³/3|₀ ^R][–cos(φ)|₀^2π] = 0.

And the z component yields

Fz	= –Σ ò0 R ds ò02π[3s2 sin(φ) + 2s4 cos2(φ)sin(φ)]
	= –3Σ [ s3/3|0 R][–cos(φ)|02π] – 2Σ [ s5/5|0 R][–cos3(φ)|02π]
	= 0



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10. 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. The magnetic field in the area is B = (5, –y, z). Determine the total magnetic force acting on the sphere.



The force on the ribbon is given by F = ò_sphere J ´ B dV . We need to express J in ijk notation, which involves determining the relationship between φ and i and j. As we saw in the previous problem, φ = –i sin(φ) + j cos(φ). As well, the magnetic field is B = i 5 – j rsin(θ)cos(φ) + k rcos(θ), where we have expressed x, y, and z in terms of r, θ , and φ . The volume element dV = r²sin(θ)dθ dφ for spherical coordinates. The cross product is



The x component of the force is thus

Fx	= ρ Ω ò0 R dr ò0πdθ ò02πdφ [r2cos(θ)sin(θ)cos(φ)]r2sin(θ)
	= ρ Ω [r5/5|0 R][(1/3)sin3(φ)|0π][sin(φ)|02π]
	= 0

The y component is

Fy	= ρ Ω ò0 R dr ò0πdθ ò02πdφ [r2cos(θ)sin(θ)sin(φ)]r2sin(θ)
	= ρ Ω [r5/5|0 R][(1/3)sin3(φ)|0π][–cos(φ)|02π]
	= 0

And the z component yields

Fz	= –ρ Ω ò0 R dr ò0πdθ ò02πdφ [r2sin2(θ)sin2(φ)–5r sin(θ)cos(φ)]r2sin(θ)
	= –ρ Ω [r5/5|0 R][–(1/3)sin2(θ)cos(θ)–(2/3)cos(θ)|0π][½φ –½cos(φ)sin(φ)|02π]
	+ 5ρ Ω [r3/3|0 R] [½θ –½cos(θ)sin(θ)|0π] [sin(φ)|02π]
	= –(4π /15)ρ Ω R5



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11. A tapered cylinder has its central axis oriented in the z direction. Its base is in the xy plane. The base has radius R, the top has radius a, and the cylinder has height h. A constant current I enters the base. The magnetic field in the area is B = (2xy, 2z+3, 5–yz). Determine the total magnetic force acting on the cylinder.



The force on the cylinder is given by F = ò_cyl J ´ B dV , where J is the volume current density. We know its direction is k since I is upwards. Now the current is defined J = dI/dA where dA is the perpendicular cross-sectional area. Clearly, at the bottom J = I/π R² and at the top J = I/π a². So we just need to know how the area depends on height z. Consider the diagram below where we have a cross-sectional view of the cylinder.



The cross-sectional area at height z is π r². Using a little geometry, we see that r' = ztan(θ)and r = R – r' = R – z tan(θ). However tan(θ)= (R – a)/h, so r = R – z(R – a)/h. Therefore J = k I / π [R – z(R – a)/h]².

In cylindrical coordinates, x = scos(φ) , y = ssin(φ), and z = z. Thus B = i 2s² cos(φ)sin(φ) + j (2z + 3) + k (5 – szsin(φ)). As well, dV = sdsdφ dz. Note that the limits of integration for z are from 0 to h, from 0 to 2π for φ , and from 0 to r = R – z(R – a)/h for s. The cross product is Jk × B



The x component of the force is thus

Fx	= (I/π ) ò0 h dz ò0 r ds ò02πdφ (2z + 3)/[R – z(R – a)/h]2
	= (I/π ) ò0 h dz ò0 r ds (2z + 3)/[R – z(R – a)/h]2[φ |02π]
	= 2I ò0 h dz (2z + 3)/[R – z(R – a)/h]2[s |0 r]
	= 2I ò0 h dz (2z + 3)r/[R – z(R – a)/h]2
	= 2I ò0 h dz (2z + 3)/[R – z(R – a)/h]
	= 2I{ ln(R/a)[3h/(R-a) + 2Rh2/(R-a)2] – 2h2/(R-a) }

The y component is

Fy	= (2I/π ) ò0 h dz ò0 r ds ò02πdφ s2 cos(φ)sin(φ)/ [R – z(R – a)/h]2
	= (2I/π ) ò0 h dz ò0 r ds s2/[R – z(R – a)/h]2 [–cos2(φ) |02π]
	= 0



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12. The force on a straight cylindrical wire of length L carrying a uniform current I in a magnetic field B₀ perpendicular to the wire is F = ILB₀. Show that you obtain exactly the same result even if the current density is not uniform but has the form J = z J(s,φ) where z is the unit vector along the cylindrical axis, s is the cylindrical radial distance from the centre of the wire, and φ is the cylindrical angle. Hint: consider what I is in this case.

For convenience let's take B = i B₀. The force on the cylinder is given by

F = ò_cyl J ´ B dV = j B₀ ò_cylJ(s,φ) sdsdφ dz = j B₀ ò₀^L dz ò₀^Rds ò₀^2π dφ sJ(s,φ) .

The integral over z may be done immediately since J doesn't depend on z. It yields the length of the wire L.

F = j B₀L ò₀^Rds ò₀^2π dφ sJ(s,φ) .

The integrals that remain are the integration of J over the cross-sectional area of the wire,

ò₀^Rds ò₀^2π dφ sJ(s,φ) = ò_A J(s,φ)dS ,

where dS = sdsdφ . This surface has unit vector z which is parallel to J. That means we can write this as

ò_A J(s,φ)dS = ò_A J • ndS = I

which by definition is the current I flowing into the wire. Thus F = IB₀L as required. That means that the currents flowing in wires could have complex patterns and we would never notice since we would get the same result as a uniform current.

Questions?mike.coombes@kwantlen.ca