Worked examples · Sheet 1

Lorentz forces on moving charges

Theory used: Lorentz force Magnetic field Field perpendicular to the page

Exercise 1

A proton moves in the plane of the page along the $x$ axis, in the positive sense, with speed $v = 3.0 \cdot 1 0^{5} m/s$ , in a uniform magnetic field of magnitude $B = 0.20 T$ directed into the page. Compute the magnetic force on the proton and its acceleration. The mass of the proton is $m_{p} = 1.67 \cdot 1 0^{- 27} kg$ .

The proton moves along the

x

axis in a field into the page; the force

F = e v \times B

is directed along the

y

axis

v = v \overset{}{^}, B = - B \hat{k}, F = e v \times B

F = e v \overset{}{^} \times (- B \hat{k}) = - e v B (\overset{}{^} \times \hat{k}) = e v B \overset{}{^}, a = \frac{e v B}{m _{p}}

F a = 1.60 \cdot 1 0^{- 19} \cdot 3.0 \cdot 1 0^{5} \cdot 0.20 N = 9.6 \cdot 1 0^{- 15} N = \frac{9.6 \cdot 1 0 ^{- 15}}{1.67 \cdot 1 0 ^{- 27}} m/ s^{2} \approx 5.7 \cdot 1 0^{12} m/ s^{2}

With the page in the $x y$ plane and the $z$ axis pointing toward the reader, a field into the page is $- B \hat{k}$ . The proton has charge $e$ , so by the Lorentz force the force is $e v \times B$ . Since $\hat{k} \times \overset{}{^} = \overset{}{^}$ , swapping the factors gives $\overset{}{^} \times \hat{k} = - \overset{}{^}$ , and the minus sign of the field makes the force directed along $+ \overset{}{^}$ , that is upward in the drawing. Velocity and field are perpendicular, therefore the magnitude of the force is the largest possible, $e v B$ . The acceleration is perpendicular to the velocity and has magnitude $5.7 \cdot 1 0^{12} m/ s^{2}$ . The value is much larger than the accelerations of macroscopic bodies, because the ratio of the charge to the mass of the proton is very large.

Exercise 2

An electron has velocity $v = (2.0 \overset{}{^} + 3.0 \overset{}{^}) \cdot 1 0^{6} m/s$ in a uniform magnetic field $B = (0.040 \overset{}{^} - 0.020 \hat{k}) T$ . Compute the magnetic force on the electron in components and its magnitude.

F = - e v \times B

v \times B = \overset{}{^} 2.0 \cdot 1 0^{6} 0.040 \overset{}{^} 3.0 \cdot 1 0^{6} 0 \hat{k} 0 - 0.020 m \cdot T/s = [3.0 \cdot 1 0^{6} \cdot (- 0.020) \overset{}{^} - 2.0 \cdot 1 0^{6} \cdot (- 0.020) \overset{}{^} - 3.0 \cdot 1 0^{6} \cdot 0.040 \hat{k}] m \cdot T/s = (- 6.0 \cdot 1 0^{4} \overset{}{^} + 4.0 \cdot 1 0^{4} \overset{}{^} - 1.2 \cdot 1 0^{5} \hat{k}) m \cdot T/s

F F = 1.60 \cdot 1 0^{- 19} \cdot (6.0 \cdot 1 0^{4} \overset{}{^} - 4.0 \cdot 1 0^{4} \overset{}{^} + 1.2 \cdot 1 0^{5} \hat{k}) N = (9.6 \overset{}{^} - 6.4 \overset{}{^} + 19.2 \hat{k}) \cdot 1 0^{- 15} N = 1.60 \cdot 1 0^{- 19} \cdot 3.6 \cdot 1 0^{9} + 1.6 \cdot 1 0^{9} + 14.4 \cdot 1 0^{9} N = 1.60 \cdot 1 0^{- 19} \cdot 1.4 \cdot 1 0^{5} N \approx 2.2 \cdot 1 0^{- 14} N

The velocity and the field are not perpendicular, so it is convenient to compute the cross product in components, with the symbolic determinant. In the determinant the term in $\overset{}{^}$ has a minus sign in front, and its minor is $v_{x} B_{z} - v_{z} B_{x} = 2.0 \cdot 1 0^{6} \cdot (- 0.020)$ ; the term in $\hat{k}$ equals $v_{x} B_{y} - v_{y} B_{x} = - 3.0 \cdot 1 0^{6} \cdot 0.040$ . The unit of the product is $m/s$ times tesla, and multiplied by a charge in coulombs it gives newtons.

The electron has charge $- e$ , so by the Lorentz force the force is the opposite of $e v \times B$ , and all the components change sign. As a check, the dot product $F \cdot v$ is proportional to $9.6 \cdot 2.0 - 6.4 \cdot 3.0 + 19.2 \cdot 0 = 0$ , so the force is perpendicular to the velocity, as it must be. The magnitude of the force is about $2.2 \cdot 1 0^{- 14} N$ .

Exercise 3

A proton moving with speed $v = 2.0 \cdot 1 0^{6} m/s$ along the $x$ axis experiences no force. When it moves with the same speed along the $y$ axis, in the positive sense, it experiences a force of magnitude $6.4 \cdot 1 0^{- 14} N$ directed in the negative sense of the $z$ axis. Determine the magnetic field, assumed uniform, and the force on the proton when its velocity, still of magnitude $v$ , lies in the $x y$ plane and forms an angle of $6 0^{\circ}$ with the $x$ axis.

F_{x} = 0 ⟹ B = B_{x} \overset{}{^}

F_{y} = e v \overset{}{^} \times B_{x} \overset{}{^} = - e v B_{x} \hat{k} = - 6.4 \cdot 1 0^{- 14} \hat{k} N ⟹ B_{x} = \frac{6.4 \cdot 1 0 ^{- 14} N}{e v}

v^{'} = v (cos 6 0^{\circ} \overset{}{^} + sin 6 0^{\circ} \overset{}{^}) ⟹ F^{'} = e v^{'} \times B_{x} \overset{}{^} = - e v B_{x} sin 6 0^{\circ} \hat{k}

B_{x} F^{'} = \frac{6.4 \cdot 1 0 ^{- 14}}{1.60 \cdot 1 0 ^{- 19} \cdot 2.0 \cdot 1 0 ^{6}} T = 0.20 T = - 6.4 \cdot 1 0^{- 14} \cdot 0.866 \hat{k} N \approx - 5.5 \cdot 1 0^{- 14} \hat{k} N

By the Lorentz force the magnetic force is zero when the velocity is parallel to the field; since the proton moving along the $x$ axis experiences no force, the field is parallel to the $x$ axis, that is $B = B_{x} \overset{}{^}$ , with $B_{x}$ positive or negative. For the velocity along the $y$ axis we use $\overset{}{^} \times \overset{}{^} = - \hat{k}$ ; the measured force is directed along $- \hat{k}$ , so $B_{x}$ is positive, and the field is $B = 0.20 \overset{}{^} T$ . The measurement with two velocities thus determines the direction, sense and magnitude of the field, as the definition of the magnetic field prescribes.

With the velocity at $6 0^{\circ}$ from the $x$ axis, the component $v cos 6 0^{\circ} \overset{}{^}$ is parallel to the field and does not contribute to the force; only the perpendicular component $v sin 6 0^{\circ} \overset{}{^}$ contributes. The force therefore has the same sense as in the previous case and its magnitude is reduced by the factor $sin 6 0^{\circ}$ , that is about $5.5 \cdot 1 0^{- 14} N$ .