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Physics Lesson 16.4.2 - Moving Trajectory of a Particle inside a Magnetic Field

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Welcome to our Physics lesson on Moving Trajectory of a Particle inside a Magnetic Field, this is the second lesson of our suite of physics lessons covering the topic of Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Moving Trajectory of a Particle inside a Magnetic Field

In the previous paragraph, we explained that magnetic force is always perpendicular to the moving direction of charges. (Do not confuse the moving direction of charges - which is usually determined by the shape of conducting wire - with the direction of wire's motion - which occurs in the direction of magnetic force. They are completely different things.)

Given this, it is clear that the magnitude of velocity (that is the speed) of moving charges placed inside a magnetic field is constant, as in this case, the velocity has no component in the direction of magnetic force. This means the kinetic energy of charged particles inside a magnetic field is constant as well (KE = mv2/2). From the law of Work-Kinetic Energy conversion, we known that when the kinetic energy is constant, no work is done by the magnetic force on moving charges (W = ΔKE = KEf - KEi = 0).

Such a situation (the displacement of charges be zero when charges are in motion) can occur only if the motion is periodic, i.e. when it repeats itself in equal time intervals. The most common periodic type of motion is the uniform circular motion. Therefore, any particle placed inside a uniform magnetic field will move in uniform circular motion, where the magnetic force will act as a centripetal force, as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force

The radius of the curved path is obtained through the magnetic-centripetal force equivalence. Thus, we have

Fmag = Fcp
Q ∙ v ∙ B = m ∙ v2/r
Q ∙ B = m ∙ v/r

Therefore, we obtain for the radius of circular path followed by a charged particle in a magnetic field:

r = m ∙ v/Q ∙ B

where m is the mass of the charged particle. Thus, if the magnetic field is uniform (B is constant), the radius r of the particle's circular path does not change as all the other quantities in the above formula are constant as well.

Given that the period of a uniform circular motion is calculated by dividing circumference by speed, we obtain for the period of rotation for a particle in circular motion inside a uniform magnetic field:

T = 2πr/v
= 2π ∙ m ∙ v/Q ∙ B/v
= 2π ∙ m/Q ∙ B

As you can see, the period of rotation is independent from the moving speed of the particle.

Example 3

Two electric charges of equal magnitudes (4 μC each) enter inside a 40 mT uniform magnetic field as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force

Calculate:

  1. The type of charge Q1 and Q2 the corresponding objects contain
  2. The masses of the two charges
  3. The speeds by which charges move inside the field
  4. The period of rotation for each charge
  5. The ratio between magnetic forces F1 / F2 produced

Take mproton = 1.67 × 1027 kg, melectron = 9.1 × 10-31 kg, e = 1.6 × 10-19 C.

Solution 3

  1. Using the Fleming's Left Hand Rule, we find out that the charge Q1 is positive and Q2 is negative. Another way to prove this is to see the turning direction of charges. Thus, when a positive charge enters inside an onto-the-page magnetic field, the magnetic force is equal to the corresponding centripetal force, producing an anticlockwise turning effect as explained earlier, while when the charge is negative, it rotates in the opposite direction (clockwise).
  2. Since all values of electric charges - whether positive or negative - are multiples of elementary charge, first we find the number n of protons or electrons contained in each charge and then, we multiply them by the mass of the corresponding elementary charges e. We have
    Q = n ∙ e ⟹ n = Q/e
    = 4 × 10-6 C/1.6 × 10-19
    = 2.5 × 1013 charges
    Since Q1 = + 4μC and Q2 = -4μC, we obtain for the masses m1 and m2 of the charged objects:
    m1 = n ∙ mp
    = 2.5 × 1013 ∙ 1.67 × 10-27 kg
    = 4.175 × 10-14 kg
    and
    m2 = n ∙ me
    = 2.5 × 1013 ∙ 9.1 × 10-31 kg
    = 2.275 × 10-17 kg
  3. The speeds of each charged object are calculated using the equation
    r = m ∙ v/Q ∙ B
    From the figure we can see that r1 = 3 cm = 3 × 10-2 m (3 units in the figure), and r2 = 1 cm = 10-2 m (1 unit in the figure). Therefore, for the positive charge Q1 we have
    v1 = r1 ∙ Q1 ∙ B/m1
    = (3 × 10-2 m) ∙ (4 × 10-6 C) ∙ (4 × 10-2 T)/4.175 × 10-14 kg
    = 11.497 × 104 m/s
    = 114 970 m/s
    and for the negative charge Q2, we have
    v2 = r2 ∙ Q2 ∙ B/m2
    = (1 × 10-2 m) ∙ (4 × 10-6 C) ∙ (4 × 10-2 T)/2.275 × 10-17 kg
    = 7.033 × 107 m/s
    = 70 330 000 m/s
  4. The period of rotation for the first charge is
    T1 = 2π ∙ m1/Q1 ∙ B
    = (2 ∙ 3.14) ∙ (4.175 × 10-14 kg)/(4 × 10-6 C) ∙ (4 × 10-2 T)
    = 1.64 × 10-6 s
    and that of the second charge is
    T2 = 2π ∙ m2/Q2 ∙ B
    = (2 ∙ 3.14) ∙ (2.275 × 10^(-17) kg)/(4 × 10-6 C) ∙ (4 × 10-2 T)
    = 8.93 × 10-10 s
  5. We can calculate the magnetic forces acting on each charge in two ways:
    1. Using the formula F = Q ∙ v ∙ B for each charge, or
    2. Using the formula of centripetal force F = m ∙ v2 / r for each charge.
    Let's use the second method in this exercise. For the first charge, we have
    F1 = m1 ∙ v12/r1
    = (4.175 × 10-14 kg) ∙ (11.497 × 104 m/s)2/(3 × 10-2 m)
    = 1.84 × 10-2 N
    and for the second charge, we have
    F2 = m2 ∙ v22/r2
    = (2.275 × 10^(-17) kg) ∙ (7.033 × 107 m/s)2/1 × 10-2 m
    = 112.5 × 10-1 N
    = 11.25 N
    These results are confirmed by means of the first method. Thus, for the first magnetic force, we have
    F1 = Q1 ∙ v1 ∙ B = (4 × 10-6 C) ∙ (11.497 × 104 m/s) ∙ (4 × 10-2 T)
    = 184 × 10-4 N
    = 1.84 × 10-2 N
    and for the second magnetic force, we have
    F2 = Q2 ∙ v2 ∙ B
    = (4 × 10-6 C) ∙ (7.033 × 107 m/s) ∙ (4 × 10-2 T)
    = 112.5 × 10-1 N
    = 11.25 N
    As you see, the results are the same in both methods.

You have reached the end of Physics lesson 16.4.2 Moving Trajectory of a Particle inside a Magnetic Field. There are 3 lessons in this physics tutorial covering Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force, you can access all the lessons from this tutorial below.

More Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force Lessons and Learning Resources

Magnetism Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial		Revision
Notes		Revision
Questions
16.4Magnetic Force on a Wire Moving Inside a Magnetic Field. Lorentz Force
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
16.4.1Magnetic Force on Moving Charges
16.4.2Moving Trajectory of a Particle inside a Magnetic Field
16.4.3Lorentz Force

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