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Questions: 1 2 3 4 5 6 7 8 

Magnetic Fields Solutions

  1. In the diagrams below, draw or indicate the direction of the magnetic force on the moving charge and calculate its magnitude.

    The magnitude of the magnetic force is given by F = |qvBsin(θ)|. Determining the direction of the force involves the following. First the velocity vector, v, and the magnetic field vector, B, define a plane. The magnetic force is perpendicular to this plane, either into or out of it. We use the Right Hand Rule to determine which. We rotate our right hand palm in the most manner which is most comfortable. The thumb of the right hand is now perpendicular to the plane and a positive charge will experience a force in the direction along the thumb. A negative charge will be anti-parallel to the direction of the thumb.

    (a) q = +5.0 μC, v = 15.0 × 103 m/s, B = 0.25 T, θ = 65°

    F = 1.699 × 10-2 N

    The plane formed by v and B is the surface of the paper. Turning your palm from v to B, your thumb points into the paper. The charge is positive, so the force is into the paper.

    (b) q = -3.0 μC, v = 6.0 × 103 m/s, B = 0.25 T, θ = 122°

    F = 3.816 × 10-3 N

    The plane formed by v and B is the surface of the paper. Turning your palm from v to B, your thumb points into the paper. The charge is negative, so the force is out the paper.

    (c) q = +5.0 μC, v = 15.0 × 103 m/s, B = 0.25 T

    Note θ = 90°, F = 1.875 × 10-2 N

    The edge of the plane formed by v and B runs along v and is the perpendicular to surface of the paper. Turning your palm from v into the paper along B, your thumb points along the paper. The charge is positive, so the force is as shown in the diagram.

    (d) q = -5.0 μC, v = 15.0 × 103 m/s, B = 0.25 T


    Note θ = 90°, F = 1.875 × 10-2 N

    The edge of the plane formed by v and B runs along v and is the perpendicular to surface of the paper. Turning your palm from v into the paper along B, your thumb points along the paper. The charge is negative, so the force is as shown in the diagram.

    (e) q = +5.0 μC, v = 15.0 × 103 m/s, B = 0.25 T


    Note θ = 90°, F = 1.875 × 10-2 N

    The edge of the plane formed by v and B runs along v and is the perpendicular to surface of the paper. Turning your palm from v out of the paper along B, your thumb points along the paper. The charge is positive, so the force is as shown in the diagram.

    (f) q = -5.0 μC, v = 15.0 × 103 m/s, B = 0.25 T

    Note θ = 90°, F = 1.875 × 10-2 N

    The edge of the plane formed by v and B runs along v and is the perpendicular to surface of the paper. Turning your palm from v out of the paper along B, your thumb points along the paper. The charge is negative, so the force is as shown in the diagram.

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  2. An electron is travelling through a uniform magnetic field of magnitude B = 1.5 × 10-3 T directed as in the diagram below. 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.

    Since the weight acts down, the magnetic force on the electron must act up to cancel. Since the velocity v is perpendicular to F and B, this means that the electron can only be moving into the page or out of the page. Using the right hand rule, a velocity crossed into the given magnetic field direction, would end up yielding a downward force for a positive charge. Since the electron has a negative charge, the electron moving into the page does give an upward force. This upward force has magnitude F = evB and must cancel the weight W = mg. Hence we have

    v = mg/eB = (9.11×10-31 kg)(9.81 m/s2)/(1.602×10-19 C)(1.50×10-3 T) = 3.7 × 10-8 m/s .

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  3. In the diagram below, a 6.00 m long wire carrying a current of 120 A is immersed in a uniform magnetic field of magnitude 0.200 T and width 3.50 m. Determine the magnetic force on the wire.

    Only the portion of the wire in the magnetic field experience the force. The magnitude of the magnetic force is given by F = ILBsin(θ). Here,

    F = (120 A)(3.50 m)(0.200 T)sin(90°) = 84.0 N .

    The line experiences a total magnetic force of 84.0 N. Note that the direction of the force would be towards the top of the paper.

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  4. A electric power line carries a current of 1400 A in a location where the earth's magnetic field is 5.0 × 10-5 T. The line makes an angle of 75° with respect to the field. Determine the magnitude of the magnetic force on a 120-m length of line.

    The magnitude of the magnetic force is given by F = ILBsin(θ). Here,

    F = (1400 A)(120 m)(5 × 10-5 T)sin(75°) = 8.11 N .

    The 120-m line experiences a total magnetic force of 8.11 N. Note that the direction of the force would be out of the paper.

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  5. The earth's magnetic field is approximately 5 × 10-5 T. If an electron is travelling 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 = qvBsin(90°) = evB. It 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 = 2πR/T = 2πRf ,

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

    f = v/(2πR) . (2)

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

    f = eB/2πm = (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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  6. 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 = qvBsin(90°) = qvB. It 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 = 2πR/T = 2πRf ,

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

    f = v/(2πR) = qB/2πm = (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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  7. 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. NA = 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 5, 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 = NAvagadro 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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  8. 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 7, 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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