Worked examples · Sheet 2

Circular motions in a uniform magnetic field

Theory used: Circular motion in a uniform magnetic field Radius, momentum and energy Work of the magnetic force

Exercise 1

An electron enters with speed $v = 4.0 \cdot 1 0^{6} m/s$ a uniform magnetic field of magnitude $B = 2.0 \cdot 1 0^{- 3} T$ , directed into the page and perpendicular to the velocity. Compute the radius of the trajectory, the period and the frequency of the motion, and determine the sense of rotation.

The electron, with negative charge, in a field into the page moves clockwise along the circle

e v B = m_{e} \frac{v ^{2}}{r} ⟹ r = \frac{m _{e} v}{e B}, T = \frac{2 π r}{v} = \frac{2 π m _{e}}{e B}, f = \frac{1}{T}

r T f = \frac{9.11 \cdot 1 0 ^{- 31} \cdot 4.0 \cdot 1 0 ^{6}}{1.60 \cdot 1 0 ^{- 19} \cdot 2.0 \cdot 1 0 ^{- 3}} m \approx 1.1 \cdot 1 0^{- 2} m = \frac{2 π \cdot 9.11 \cdot 1 0 ^{- 31}}{1.60 \cdot 1 0 ^{- 19} \cdot 2.0 \cdot 1 0 ^{- 3}} s \approx 1.8 \cdot 1 0^{- 8} s = \frac{1}{1.79 \cdot 1 0 ^{- 8} s} \approx 5.6 \cdot 1 0^{7} Hz

The only force on the electron is the magnetic force, which has constant magnitude $e v B$ and is perpendicular to the velocity; by circular motion in a uniform magnetic field the electron moves along a circle, and the magnetic force provides the centripetal acceleration. From this equality follow the radius, about $1.1 cm$ , and the period, which does not depend on the speed.

For the sense we consider the lowest point of the trajectory, where the force must be directed upward, that is toward the center. With the field into the page, $B = - B \hat{k}$ , and the velocity to the left, $v = - v \overset{}{^}$ , we have $v \times B = v B (\overset{}{^} \times \hat{k}) = - v B \overset{}{^}$ , directed downward; the charge is negative, so the force $- e v \times B$ is directed upward. The electron therefore rotates clockwise.

Exercise 2

A proton, initially at rest, is accelerated by the potential difference $Δ V = 5.0 kV$ and enters a uniform magnetic field $B = 0.25 T$ perpendicular to its velocity. Compute the speed of the proton, the radius of the trajectory and the time it takes to travel half a circle. The mass of the proton is $m_{p} = 1.67 \cdot 1 0^{- 27} kg$ .

e Δ V = \frac{1}{2} m_{p} v^{2} ⟹ v = \frac{2 e Δ V}{m _{p}}, r = \frac{m _{p} v}{e B}, t = \frac{T}{2} = \frac{π m _{p}}{e B}

v r t = \frac{2 \cdot 1.60 \cdot 1 0 ^{- 19} \cdot 5.0 \cdot 1 0 ^{3}}{1.67 \cdot 1 0 ^{- 27}} m/s \approx 9.8 \cdot 1 0^{5} m/s = \frac{1.67 \cdot 1 0 ^{- 27} \cdot 9.79 \cdot 1 0 ^{5}}{1.60 \cdot 1 0 ^{- 19} \cdot 0.25} m \approx 4.1 \cdot 1 0^{- 2} m = \frac{π \cdot 1.67 \cdot 1 0 ^{- 27}}{1.60 \cdot 1 0 ^{- 19} \cdot 0.25} s \approx 1.3 \cdot 1 0^{- 7} s

During the acceleration the electric field does the work $e Δ V$ on the proton, and the proton starts from rest; by the work-energy theorem this work equals the final kinetic energy, $5.0 keV$ , from which the speed follows. In the magnetic field the velocity is perpendicular to $B$ , so by circular motion in a uniform magnetic field the proton describes a circle of radius about $4.1 cm$ .

By the work of the magnetic force the speed remains $9.8 \cdot 1 0^{5} m/s$ throughout the motion, and half a circle takes half the period. As a check, $π r$ divided by $v$ gives $\frac{π \cdot 4.09 \cdot 1 0 ^{- 2}}{9.79 \cdot 1 0 ^{5}} s \approx 1.3 \cdot 1 0^{- 7} s$ , the same value.

Exercise 3

A proton and an alpha particle, with charge $2 e$ and mass $m_{α} \approx 4 m_{p}$ , have the same kinetic energy $K = 1.0 MeV$ and move in a uniform magnetic field $B = 0.80 T$ , perpendicular to the field. Compute the radius of the trajectory and the period of each particle. The mass of the proton is $m_{p} = 1.67 \cdot 1 0^{- 27} kg$ .

r = \frac{2 m K}{∣ q ∣ B}, T = \frac{2 π m}{∣ q ∣ B}

r_{α} = \frac{2 \cdot 4 m _{p} K}{2 e B} = \frac{2 m _{p} K}{e B} = r_{p}, T_{α} = \frac{2 π \cdot 4 m _{p}}{2 e B} = 2 T_{p}

K r_{p} T_{p} = 1.0 \cdot 1 0^{6} \cdot 1.60 \cdot 1 0^{- 19} J = 1.6 \cdot 1 0^{- 13} J = \frac{2 \cdot 1.67 \cdot 1 0 ^{- 27} \cdot 1.6 \cdot 1 0 ^{- 13}}{1.60 \cdot 1 0 ^{- 19} \cdot 0.80} m = \frac{2.31 \cdot 1 0 ^{- 20}}{1.28 \cdot 1 0 ^{- 19}} m \approx 0.18 m = \frac{2 π \cdot 1.67 \cdot 1 0 ^{- 27}}{1.60 \cdot 1 0 ^{- 19} \cdot 0.80} s \approx 8.2 \cdot 1 0^{- 8} s, T_{α} \approx 1.6 \cdot 1 0^{- 7} s

The kinetic energy is given in electronvolts and must be converted to joules. The momentum is expressed through the kinetic energy, $m v = 2 m K$ , and by the relation between radius, momentum and energy we obtain the formula for the radius as a function of $K$ . For the alpha particle the mass is four times larger and the charge is double; the numerator doubles, because it contains the square root of the mass, and the denominator doubles, so the two radii are equal, about $18 cm$ .

The period is proportional to the ratio $\frac{m}{∣ q ∣}$ , which for the alpha particle is double that of the proton; therefore the alpha particle takes twice as long to travel the same circle, consistently with the fact that, for equal energy, it is twice as slow. The speed of the proton is $\frac{2 K}{m _{p}} \approx 1.4 \cdot 1 0^{7} m/s$ , about $5 %$ of $c$ , so the nonrelativistic treatment is adequate.