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19.2Motion in a magnetic field

We just saw that the path of a charged particle in a uniform electric field is a straight line or a parabola. What is the shape of the path of a charged particle in a uniform magnetic field?

In the simple case where the velocity of the charged particle is parallel or anti-parallel to that of the uniform magnetic field, the magnetic force is zero and so the particle will move along a straight line, Figure 19.14.

Figure 19.14: When the velocity is parallel or anti-parallel to the magnetic field the force is zero.

Figure 19.15 shows a positive charge in a uniform magnetic field that is directed into the plane of the paper. The particle has a velocity v at the instant shown. The velocity is at right angles to the magnetic field. The magnetic force is at right angles to both the velocity and the magnetic field so the particle will follow a circular path; the centripetal force is provided by the magnetic force.

Figure 19.15: A charge in a magnetic field moves in a circle.

Consider a charge q moving with speed v at right angles to a magnetic field B. The force on the charge is F = qvB at right angles to the velocity. The charge moves in a circle of radius R, and so by Newton’s second law:

q v B=m v 2 R

Rearranging, we get:

R=m v q B

Very massive or very fast charges will move in large circles; large charges and large magnetic fields will result in small circles. The time T to make one full revolution in a magnetic field is found from:

T=2 π R v

T=2 π v m v q B

T=2 π m q B

This shows that T is independent of the speed. This is an important result in experimental particle physics and forms the basis for an accelerator called the cyclotron.

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CHECK YOURSELF 3

A particle of mass m and charge q moves on a circular path in a region of magnetic field B at right angles to the velocity. What is the average magnetic force experienced by the particle during half a revolution?

Suppose, however, that the charged particle loses energy as it moves along the circular path. (This could be because the particle is colliding with other particles or it radiates electromagnetic waves.) The speed will decrease, and from the formula for the radius we see that the radius will also decrease. This means that the circular path will become a spiral, Figure 19.16.

Figure 19.16: A negatively charged particle spiraling inwards in a region of magnetic field directed into the page.

Worked Example 19.4

Figure 19.17 shows a charged particle entering a region of magnetic field that is directed into the plane of the page.

The path of the particle is a quarter circle. The speed of the particle is 5.2 × 106 m s−1.

Figure 19.17: For worked example 19.4.

aJustify why the charge is positive.

bThe particle is in fact a proton with mass 1.67 × 10−27 kg and charge 1.6 × 10−19 C. The magnetic flux density is 0.25 T. Calculate the radius of the proton’s circular path.

cCalculate the time the proton spends in the region of magnetic field.

Answer

aThe force must be directed towards the centre of the circle. The field is into the page so by the right-hand force rule the charge must be positive.

bFrom q v B=m v 2 R we deduce that R=m v q B. Thus:

R=1.67×10−27×5.2×10 6 1.6×10−19×0.25

R=0.217≈0.22 m

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cThe path is a quarter of a circle of radius R, so the length of the path is:

2 π R 4=2 π×0.217 4=0.34 m

The time in the field is therefore:

0.34 5.2×10 6=6.6×10−8 s

Worked Example 19.5

Figure 19.18 shows the path of a charged particle. The particle goes through a thin metallic foil.

State and explain the direction of motion of the particle and the sign of its charge.

Figure 19.18: For worked example 19.5.

Answer

The path consists of two circular arcs of different radius. The radius gets smaller because the particle loses energy as it passes through the foil. Therefore the direction of motion is counterclockwise. Since the field is directed into the plane of the page the charge must be positive by the right-hand force rule so that the force is directed towards the centre of the arcs.

Work done and magnetic forces

Since the magnetic force is always normal to the velocity of the charge, it follows that it cannot do any work. The big magnets in particle accelerators are used only to deflect particles, not to increase the particles’ kinetic energy (that job is done by electric fields). This means the kinetic energy of the particle cannot change while in the region of magnetic field.

Exam tip

W=F s cos θ

For the magnetic force, θ = 90° giving W = 0.

CHECK YOURSELF 4

Why does the proton in worked example 19.4 exit the region of magnetic field with the same speed as that at the entry point?

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Motion in crossed electric and magnetic fields

Finally, we examine the motion of a charged particle in a region where an electric field and a magnetic field are both present. We will deal with the case in which the electric field is at right angles to the magnetic field. Figure 19.19 shows a positively charged particle entering, with speed v, a region of crossed E and B fields.

Figure 19.19: a A positively charged particle entering a region of crossed electric and magnetic fields. b A free-body diagram showing the forces on the particle.

The particle will be acted upon by an electric force F e = q E directed along the electric field and a magnetic force F m = q v B directed opposite to the electric force. If these forces are equal, the particle will continue moving on a straight line undef lected from its original path. This means that

q E=q v B⇒E=v B.

The interesting fact in this case is that the charge has cancelled out; so if the condition E = v B holds, any charge of any value and any sign will continue undeflected through the crossed fields as long as it enters with velocity v.

If the particle enters with speed greater than v, the magnetic force is larger and the particle will move upwards; it will move downwards if the speed is less than v.

Links

The study of the motion of charged particles in electric and magnetic fields has made possible the construction of particle accelerators. For example, in synchrotrons like the Large Hadron Collider at CERN, a beam of protons is accelerated to speeds very close to the speed of light by electric fields. Powerful magnets force the protons to move along a circular track. As the speed of the protons increases the strength of the magnets must also increase in order to keep the protons on the same circular track, thus requiring the development of sophisticated superconducting electromagnets. Because the speeds involved are close to that of light, relativity (Chapter 6) must be taken into account in constructing these machines. The accelerated particles collide with another beam of accelerated particles travelling in the opposite direction and during the collision energy gets converted into matter, producing new particles. We have obtained much of our knowledge about the structure of matter (Chapters 21 through 25) through the results of accelerator experiments.

TEST YOUR UNDERSTANDING

7How can a charged particle follow a straight line path in a uniform magnetic field?

8An electron with speed 3.2 x 107 m s−1 enters a region of uniform magnetic field 0.025 T. The direction of velocity is at right angles to the field. Calculate the radius of the circular path.

9Protons move in a circular path of radius 0.75 m with speed 3.0 × 107 m s−1.

What is the magnetic field required?

10A proton moves in a circular path of radius 12 cm in a region of magnetic field of flux density 0.35 T. The field is at right angles to the velocity. What is the speed of the proton?

11 aA proton of mass m p moves in a circular path in a region of magnetic field B directed normally to its velocity. Show that the number of revolutions per second is e B 2 π m p.

bThe proton is replaced by an alpha particle (mass 4m p, charge +2e). How does the answer to a change, if at all?

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12An electron is accelerated from rest by a potential difference 250 V. It then enters a region of uniform magnetic field where it is bent into a semicircular path (Figure 19.20). The time to complete the semicircle is 1.5 ns.

Figure 19.20: For question 12.

aCalculate the speed of the electron as it enters the region of magnetic field.

bHence, calculate the radius of the semicircle.

cDetermine the magnetic field.

13An electron is accelerated from rest by a potential difference V. It then enters a region of magnetic field of strength B as shown in Figure 19.21. The electron follows a semicircular path of radius R.

Figure 19.21: For question 13.

aShow that the charge-to-mass ratio of the electron is given by e m=2 V B 2 R 2.

bIn one experiment V = 490 V, B = 1.2 mT and R = 6.1 cm. Calculate the charge-to-mass ratio of the electron.

cUsing a value 1.6 × 10−19 C for the charge, estimate the mass of the electron from these data.

14A singly ionised beam of ions enters a region of magnetic field with speed 1.3 × 106 m s−1. The beam contains two types of different atoms of the same element. The two types of atoms differ in the number of neutrons they contain.

The magnetic flux density is 0.21 T. The ions are bent into two circular paths that differ in radius by 0.13 m (Figure 19.22).

Figure 19.22: For question 14.

What is the difference in mass of the two types of atoms in the beam?

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15Figure 19.23 shows two parallel plates. The electric field is directed from top to bottom and has magnitude 2.4 × 103 NC−1. The shaded region is a region of magnetic field normal to the page.

Figure 19.23: For question 15.

aDeduce the magnetic field magnitude and direction so that an electron experiences zero net force when shot through the plates with a speed of 2.0 × 105 m s−1.

bSuggest whether a proton shot with the same speed through the plates experiences zero net force.

cWould an alpha particle (mass 4 times that of a proton, charge double that of a proton) be undeflected if it entered with the same speed as in a?

dThe electron’s speed is doubled. Suggest whether the electron would still be undeflected for the same magnetic field found in a.

16An electron enters a region of uniform vertical magnetic field, B (Figure 19.24). The initial velocity v of the electron makes an angle θ with the horizontal.

Figure 19.24: For question 16.

aBy reference to the horizontal and vertical components of the velocity of the electron, explain why the path of the electron will be a spiral.

bShow that the radius of the spiral is R=m v cos θ e B.

cDerive an expression for the pitch, p, of the spiral. The pitch is the vertical distance travelled in a time equal to the time taken for one full revolution.