The Timpson Lyceum

1. (2 pts ••) A proton moves with speed $v=2.5\times10^{5}\,\text{m/s}$ through a uniform magnetic field of magnitude $B=0.85\,\text{T}$ . Its velocity makes an angle of $30^\circ$ with the field.

(a) Find the magnitude of the magnetic force on the proton.

(b) The velocity points in the $+x$ direction and the field lies in the $xy$ -plane, tilted $30^\circ$ above the $x$ -axis toward $+y$ . In which direction does the force on the proton point? Justify using the right-hand rule.

(c) If the particle were an electron with the same velocity, how would the force direction change?

Solution

1.

(a) $F=qvB\sin\theta=(1.60\times10^{-19})(2.5\times10^{5})(0.85)\sin30^\circ\approx 1.7\times10^{-14}\,\text{N}$ .

(b) With $\mathbf{v}=v\,\hat{\mathbf{x}}$ and $\mathbf{B}$ in the $xy$ -plane (components along $+x$ and $+y$ ), $\mathbf{v}\times\mathbf{B}$ has only the component from $\hat{\mathbf{x}}\times\hat{\mathbf{y}}=\hat{\mathbf{z}}$ . So $\mathbf{v}\times\mathbf{B}$ points in $+\hat{\mathbf{z}}$ (out of the plane), and since $q>0$ the force $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$ also points in $+\hat{\mathbf{z}}$ (out of the page).

(c) An electron has $q<0$ , so the force reverses: it points in $-\hat{\mathbf{z}}$ (into the page). The magnitude is the same.

2. (3 pts •••) A charged particle enters a region of uniform magnetic field with its velocity perpendicular to the field and travels in a circle.

(a) Starting from $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$ and Newton's second law, derive the expression for the radius $r=mv/qB$ of the circular path.

(b) An electron moves at $v=3.0\times10^{6}\,\text{m/s}$ perpendicular to a field $B=0.45\,\text{T}$ . Find the radius of its orbit.

(c) Find the period of the motion and the cyclotron frequency $f=qB/2\pi m$ . Confirm numerically that these do not depend on the speed.

Solution

2.

(a) When $\mathbf{v}\perp\mathbf{B}$ , the force $F=qvB$ is always perpendicular to $\mathbf{v}$ , so it does no work and only changes direction — uniform circular motion. Setting the magnetic force equal to the required centripetal force:

$qvB=\frac{mv^2}{r}\quad\Rightarrow\quad r=\frac{mv}{qB}.$

(b) $r=\dfrac{m_e v}{eB}=\dfrac{(9.11\times10^{-31})(3.0\times10^{6})}{(1.60\times10^{-19})(0.45)}\approx 3.8\times10^{-5}\,\text{m}$ (about $38\,\mu\text{m}$ ).

(c) Period $T=\dfrac{2\pi m}{qB}=\dfrac{2\pi(9.11\times10^{-31})}{(1.60\times10^{-19})(0.45)}\approx 7.9\times10^{-11}\,\text{s}$ .

Cyclotron frequency $f=\dfrac{qB}{2\pi m}=\dfrac{1}{T}\approx 1.3\times10^{10}\,\text{Hz}$ ( $\approx13\,\text{GHz}$ ). Neither $T$ nor $f$ contains $v$ : a faster particle simply travels a proportionally larger circle in the same time.

3. (2 pts ••) A velocity selector uses crossed electric and magnetic fields so that only particles with one speed pass straight through undeflected. The magnetic field is $B=0.12\,\text{T}$ , and singly charged ions of speed $v=4.0\times10^{5}\,\text{m/s}$ are to pass through.

(a) What electric field magnitude is required, and how must $\mathbf{E}$ and $\mathbf{B}$ be oriented relative to $\mathbf{v}$ ?

(b) The field is produced by two parallel plates separated by $1.5\,\text{cm}$ . What voltage must be applied across them?

(c) Explain what happens to an ion moving faster than the selected speed.

Solution

3.

(a) For straight-through motion the electric and magnetic forces cancel: $qE=qvB$ , so $E=vB=(4.0\times10^{5})(0.12)\approx 4.8\times10^{4}\,\text{V/m}$ . The fields $\mathbf{E}$ , $\mathbf{B}$ , and $\mathbf{v}$ must be mutually perpendicular, with $\mathbf{E}$ oriented so the electric force opposes the magnetic force $q\mathbf{v}\times\mathbf{B}$ .

(b) $V=Ed=(4.8\times10^{4})(0.015)\approx 720\,\text{V}$ .

(c) A faster ion has a larger magnetic force ( $qvB$ ) than electric force ( $qE$ ), so the magnetic force wins and it is deflected out of the straight path (toward the direction of the net magnetic force). Only $v=E/B$ passes undeflected.

4. (2 pts ••) A straight wire of length $L=0.35\,\text{m}$ carries a current $I=8.5\,\text{A}$ in a uniform magnetic field $B=0.60\,\text{T}$ .

(a) Find the force on the wire when the current is perpendicular to $\mathbf{B}$ .

(b) Find the force when the wire makes an angle of $55^\circ$ with $\mathbf{B}$ .

(c) At what angle is the force zero?

Solution

4.

(a) $F=BIL\sin90^\circ=(0.60)(8.5)(0.35)\approx 1.8\,\text{N}$ .

(b) $F=BIL\sin55^\circ=(0.60)(8.5)(0.35)\sin55^\circ\approx 1.5\,\text{N}$ .

(c) The force is zero when $\sin\theta=0$ , i.e. when the current is parallel (or antiparallel) to $\mathbf{B}$ ( $\theta=0^\circ$ or $180^\circ$ ).

5. (2 pts ••) A flat rectangular coil of $N=50$ turns, dimensions $4.0\,\text{cm}\times6.0\,\text{cm}$ , carries a current $I=1.2\,\text{A}$ in a uniform field $B=0.30\,\text{T}$ .

(a) Find the magnetic dipole moment $\mu=NIA$ of the coil.

(b) Find the maximum torque the field can exert on the coil.

(c) Find the work required to rotate the coil from the orientation of minimum potential energy (moment aligned with $\mathbf{B}$ ) to the orientation where the moment is antiparallel to $\mathbf{B}$ .

Solution

5.

(a) $A=(0.040)(0.060)=2.4\times10^{-3}\,\text{m}^2$ , so $\mu=NIA=(50)(1.2)(2.4\times10^{-3})\approx 0.14\,\text{A}\cdot\text{m}^2$ .

(b) $\tau_{\max}=\mu B\sin90^\circ=(0.144)(0.30)\approx 0.043\,\text{N}\cdot\text{m}$ .

(c) Using $U=-\mu B\cos\theta$ : aligned ( $\theta=0$ ) gives $U_i=-\mu B$ ; antiparallel ( $\theta=180^\circ$ ) gives $U_f=+\mu B$ . The work done against the field is

$W=U_f-U_i=2\mu B=2(0.144)(0.30)\approx 0.086\,\text{J}.$

6. (2 pts ••) A copper ribbon $0.50\,\text{mm}$ thick and $2.0\,\text{cm}$ wide carries a current $I=25\,\text{A}$ perpendicular to a magnetic field $B=0.80\,\text{T}$ directed through the flat face of the ribbon. The carrier density of copper is $n=8.5\times10^{28}\,\text{m}^{-3}$ .

(a) Find the Hall voltage measured across the width of the ribbon.

(b) Explain how the sign of the Hall voltage reveals whether the charge carriers are positive or negative.

Solution

6.

(a) The Hall field balances the magnetic force on the carriers: $qE_H=qv_dB$ , and $I=nqv_dA=nqv_d(wt)$ gives $v_d=I/(nqwt)$ . The Hall voltage across the width $w$ is $V_H=E_H w=v_dBw=\dfrac{IB}{nqt}$ :

$V_H=\frac{(25)(0.80)}{(8.5\times10^{28})(1.60\times10^{-19})(0.50\times10^{-3})}\approx 2.9\times10^{-6}\,\text{V}$

(about $2.9\,\mu\text{V}$ ; note $V_H$ is independent of the ribbon width).

(b) The magnetic force $q\mathbf{v}\times\mathbf{B}$ pushes carriers to one edge. Positive carriers drifting in the current direction and negative carriers drifting opposite to the current are pushed to the same edge, but their opposite charge makes that edge acquire opposite sign. So the polarity (sign) of the measured Hall voltage tells you the sign of the mobile charge — this is how the Hall effect established that conduction in metals is by negative electrons.

7. (1 pt •) An electron enters a uniform field $B=0.30\,\text{T}$ with speed $v=5.0\times10^{5}\,\text{m/s}$ , its velocity making an angle of $25^\circ$ with the field, so it follows a helix.

(a) Find the radius of the helix.

(b) Find the pitch (the distance advanced along the field per revolution).

Solution

7.

(a) Only the perpendicular velocity component curves the path: $v_\perp=v\sin25^\circ$ , so

$r=\frac{m_e v_\perp}{eB}=\frac{(9.11\times10^{-31})(5.0\times10^{5})\sin25^\circ}{(1.60\times10^{-19})(0.30)}\approx 4.0\times10^{-6}\,\text{m}.$

(b) The parallel component $v_\parallel=v\cos25^\circ$ advances the particle during one period $T=2\pi m_e/eB\approx 1.19\times10^{-10}\,\text{s}$ :

$\text{pitch}=v_\parallel T=(5.0\times10^{5}\cos25^\circ)(1.19\times10^{-10})\approx 5.4\times10^{-5}\,\text{m}.$

Extra Credit (3 pts •••)

A particle of charge $q$ and mass $m$ starts from rest at the origin in a region of crossed static fields: a uniform electric field $\mathbf{E}=E\,\hat{\mathbf{y}}$ and a uniform magnetic field $\mathbf{B}=B\,\hat{\mathbf{z}}$ .

(a) Write the equations of motion for $v_x(t)$ and $v_y(t)$ from $\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ , and show they are coupled first-order equations.

(b) Solve them with the initial condition $\mathbf{v}(0)=0$ . (Hint: define the cyclotron frequency $\omega=qB/m$ and look for the steady drift plus an oscillation.)

(c) Integrate to find $x(t)$ and $y(t)$ , and show the trajectory is a cycloid with a constant drift velocity $v_d=E/B$ in the $+x$ direction. Sketch the path and give the maximum displacement of the particle in the $y$ -direction.

Solution

Extra Credit

(a) With $\mathbf{E}=E\,\hat{\mathbf{y}}$ , $\mathbf{B}=B\,\hat{\mathbf{z}}$ , and $\mathbf{v}=(v_x,v_y,0)$ ,

$\mathbf{v}\times\mathbf{B}=(v_x\hat{\mathbf{x}}+v_y\hat{\mathbf{y}})\times B\hat{\mathbf{z}}=v_xB(\hat{\mathbf{x}}\times\hat{\mathbf{z}})+v_yB(\hat{\mathbf{y}}\times\hat{\mathbf{z}})=-v_xB\,\hat{\mathbf{y}}+v_yB\,\hat{\mathbf{x}}.$

Newton's second law $m\dot{\mathbf{v}}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ gives, component by component (defining $\omega=qB/m$ ):

$\dot v_x=\omega v_y,\qquad \dot v_y=\frac{qE}{m}-\omega v_x.$

These are coupled: $\dot v_x$ depends on $v_y$ and $\dot v_y$ depends on $v_x$ .

(b) Differentiate the first equation and substitute the second:

$\ddot v_x=\omega\dot v_y=\omega\!\left(\frac{qE}{m}-\omega v_x\right)=\omega\frac{qE}{m}-\omega^2 v_x.$

This is simple harmonic motion about the steady value $v_x=\dfrac{qE}{m\omega}=\dfrac{E}{B}\equiv v_d$ . The general solution is $v_x=v_d+A\cos\omega t+C\sin\omega t$ . Applying $v_x(0)=0$ gives $A=-v_d$ ; from $\dot v_x(0)=\omega v_y(0)=0$ we get $C=0$ . Hence

$v_x(t)=v_d\,(1-\cos\omega t),\qquad v_y(t)=\frac{\dot v_x}{\omega}=v_d\sin\omega t.$

(Check: $v_y(0)=0$ as required, and the drift $v_d=E/B$ is the crossed-field drift speed.)

(c) Integrate with $x(0)=y(0)=0$ :

$x(t)=v_d\!\left(t-\frac{\sin\omega t}{\omega}\right),\qquad y(t)=\frac{v_d}{\omega}(1-\cos\omega t).$

These are the parametric equations of a cycloid — the curve traced by a point on a wheel of radius $R=v_d/\omega$ rolling along the $x$ -axis at speed $v_d$ . The motion is a uniform drift $v_d=E/B$ in the $+x$ direction (independent of the particle's charge sign and mass) superposed on a circular cyclotron motion.

The particle repeatedly returns to $y=0$ (each time $\omega t=2\pi n$ ) and reaches its maximum $y$ when $\cos\omega t=-1$ :

$y_{\max}=\frac{2v_d}{\omega}=\frac{2E/B}{qB/m}=\frac{2mE}{qB^2}.$

The path looks like the humped arches of a rolling-wheel curve, marching steadily in $+x$ .

Chapter 27 — Magnetic Field & Magnetic Forces

Chapter 27 — Magnetic Field & Magnetic Forces

Extra Credit (3 pts •••)

Extra Credit