THE UNIVERSITY OF BRITISH COLUMBIA
 
Physics 108 Assignment # 7 SOLUTIONS:
 
THE MAGNETIC FIELD
 
Wed. 23 Feb. 2005 - finish by Wed. 2 Mar.

1. FRICTION vs. THE LORENTZ FORCE: A 2-kg copper rod rests on two horizontal rails 2 m apart and carries a current of 100 A from one rail to the other. The coefficient of static friction between the rod and the rails is $\mu_s = 0.5$. What is the smallest magnetic field (not necessarily vertical) that would cause the bar to slide?

   ANSWER: First let's calculate the minimum required magnetic force: the usual FBD gives $N = mg - F\sin\theta$ and $F\cos\theta = \mu N$ just before slipping. Solving for $F$ yields $F = \mu mg /[\cos\theta + \mu\sin\theta]$. Differentiating with respect to $\theta$ gives $dF/d\theta = \mu mg (\sin\theta - \mu\cos\theta)/(\cos\theta + \mu\sin\theta)^2$, which is zero (the condition for an extremum) when $\sin\theta = \mu\cos\theta$ or when $\tan\theta = \mu = 0.5\Longrightarrow \theta = 0.4636$. Thus $F = (0.5\times 2\times 9.81)/(0.8944+ 0.5\times 0.4472) = 8.774$ N. Since the current is out of the page, we want a magnetic field $\Vec{B}$ in the plane of the page at right angles to the desired force, i.e. making an angle ${\pi\over2}+\theta = 2.034$ with the horizontal $x$ direction. The magnitude of $B_{\rm min}$ is given by $F = ILB_{\rm min}$ with $L = 2$ m, namely $B_{\rm min} = F/IL = 8.774/(100\times 2)$ or  .

2. CYCLOTRONS: (Neglect any relativistic effects.) Suppose that we want to build a small cyclotron for protons using a magnet with a uniform field over a region 1 m in radius such that the protons reach a maximum kinetic energy of 20 MeV at the outer radius of the magnet.
   $(a)$ What magnetic field must the magnet produce?

   ANSWER: In each case we have $\ds{m{v^2 \over r} = qvB \Longrightarrow mv \equiv p = qBr}$ and $\ds{\omega = {v \over r} = {qB \over m}}$. In this case $q = e = 1.602 \times 10^{-19}$ C and $m = m_p = 1.67 \times 10^{-27}$ kg. At $K = 20$ MeV $= 20\times 10^6 \times 1.602 \times 10^{-19} = 3.204\times 10^{-12}$ J, $p = \sqrt{2 m K} = 1.035\times 10^{-19}$ kg-m/s, so when $r = 1$ m, $B = p/qr$ gives  .
   $(b)$ At what frequency must the "dee" voltage oscillate?

   ANSWER: $\omega = qB/m = 6.19\times 10^7$ s$^{-1}$ and  .
   Now suppose we want to build a cyclotron to accelerate electrons without a magnet, using the Earth's magnetic field (assume $B = 5 \times 10^{-5}$ T) to keep the electrons moving in circles.
   $(c)$ What is the radius of the electron orbit at 100 eV?

   ANSWER: Now $m = m_e = 9.11 \times 10^{-31}$ kg, $\vert q\vert = e$, $B = 5 \times 10^{-5}$ T and $K = 100$ eV $= 100\times 1.602 \times 10^{-19} = 1.602\times 10^{-17}$ J $\Longrightarrow p = \sqrt{2 m K} = 5.403\times 10^{-24}$ kg-m/s giving  .
   $(d)$ What is the frequency (in Hz) of the RF electric field we must supply to the cyclotron "dees?"

   ANSWER: $\omega = qB/m = 8.794 \times 10^6$ s$^{-1}$ and  .

3. HOLLOW CYLINDRICAL CONDUCTOR: A thick-walled hollow conducting cylinder carries a uniformly distributed current $I$. The (centred) hole in the middle has a radius of $R$ and the outer radius of the conductor is $2R$. Derive an expression for the strength of the magnetic field $B$ as a function of radial distance $r$ from the cylinder axis, in the range from $r = R$ to $r = 2R$; then plot (i.e. sketch, showing axis labels, scales and values at key points) $B(r)$ in the range from $r = 0$ to $r = 4R$.
   
   
   

   ANSWER: Throughout this problem, symmetry demands that the magnetic field circulate around the central axis according to the right hand rule and have the same magnitude $B(r)$ everywhere on an Ampèrian loop of radius $r$ centred on the axis. Thus $\ds{\oint \Vec{B} \cdot d\Vec{s} = 2\pi r B = \mu_0 I_{\rm encl} }$ for every such loop and all there is to the calculation is to determine the current $I_{\rm encl}$ enclosed by the loop for each $r$. When $r<R$, no current is enclosed so $B(r<R) = 0$. For $r>2R$, all the current is enclosed so $\ds{ B(r>2R) = {\mu_0 I \over 2\pi r} }$ just as for a wire along the axis. The only nontrivial case is for $R<r<2R$, where the enclosed current is given by $\ds{ {I_{\rm encl} \over I } = {\pi r^2 - \pi R^2 \over \pi(2R)^2 - \pi R^2 } }$ - i.e. the fraction of current is the fraction of cross-sectional area.

   

   This gives  . At $r = 2R$, $\ds{ B(2R) \equiv B_0 = {\mu_0 I \over 4\pi R} }$. (See sketch.)

Challenge Problems:

1. MOTION OF AN ELECTRON IN A MAGNETIC FIELD: An electron has a kinetic energy of 400 eV as it moves through a region containing a uniform magnetic field $\Vec{B} = B \kH$ of magnitude $B = 4\times 10^{-4}$ T. At $t=0$ it is at the origin of coordinates $(x=0, y=0, z=0)$ and has velocity components $v_y = 0$ and $v_x = v_z > 0$. Find the position of the electron ($x$, $y$ and $z$) 10 ns later. [1 ns = $10^{-9}$ s] ANSWER: The kinetic energy is $K = {1\over2}mv^2 = 400\times 1.602 \times 10^{-19} = 6.41\times 10^{-17}$ J and $m = 9.11\times10^{-31}$ kg, so $v = \sqrt{2K/m} = 1.186\times 10^7$ m/s. Since $\Vec{v} = v_x \iH + v_z \kH$ and $v_x \equiv v_\perp = v_z \equiv v_\parallel$, we have $v_\parallel = v_\perp = v/\sqrt{2} = 0.8388\times 10^7$ m/s. Now, $\Vec{B} = B \kH$ so $v_\parallel$ is parallel to $\Vec{B}$ and therefore continues unaffected by the magnetic field, while $v_\perp$ bends in a circle of radius $r = mv_\perp/qB = 0.1192$ m. The electron is negatively charged, so it bends to the left in an "up" field (you can verify this from the right-hand rule for $\Vec{F} = q\Vec{v} \times \Vec{B}$) and so the path is a spiral with constant radius and pitch, as shown in the figure. The frequency of the orbit of $v_\perp$ is given by $\omega = v_\perp/r = qB/m = 0.7035\times 10^8$ s$^{-1}$. At $t = 10 \times 10^{-9}$ s the angle swept out by the spiral since $t=0$ is $\theta = \omega t = 0.7035$ radians. This gives $x = r \sin \theta$ and $y = r(1 - \cos \theta)$ the values and . Meanwhile .
   

2. FORCE ON A CURRENT-CARRYING CONDUCTOR: A long, rigid conductor, lying along the $x$ axis, carries a current of 6 A in the $- \iH$ direction. A magnetic field $\Vec{B} = 3.0 \iH + 6x^2 \jH$ (with $x$ in m and $B$ in mT) is present. Calculate the vector force on the 3-m segment of the conductor that lies between $x = 1.0$ m and $x = 4$ m.

   ANSWER: In general $d\Vec{F} = I d\Vec{\ell} \times \Vec{B}$. In this case (suppressing the $SI$ units) we have $I d\Vec{\ell} = - 6\iH dx$ so $d\Vec{F} = dx (- 6\iH) \times (3.0 \iH + 6x^2 \jH) \times 10^{-3} = dx (- 0.036x^2) \kH$ (remember, $\iH \times \iH = 0$) which we must then integrate over the specified range to get $\ds{ \Vec{F} = - 0.036\kH \int_{1.0}^{4} x^2 \, dx = -\left( 0.036\over 3 \right) \kH \left[ x^3 \right]_{1.0}^{4} = - 0.012\kH (64- 1) }$ or  .

3. AMPÈRE'S LAW: A wire carrying a current of 2001 A coming out of the page, as shown, emerges from the centre of the square ABCD whose side is 3 m in length.   $(a)$ Using AMPÈRE'S LAW, find the average value along AB of the magnetic field component parallel to AB.   $(b)$ Find the magnitude and direction of the magnetic field at the midpoint of the line AB.
   

   ANSWER:   $(a)$ The magnetic field does not point directly along the Ampèrian loop everywhere, nor is it constant in magnitude along the loop; but we are only interested in the average component parallel to AB. By symmetry, whatever $\Vec{B}$ does along one side it does along the other three sides as well, so $\int_{\sf A}^{\sf B} \Vec{B} \cdot d\Vec{s} \equiv (3\; \hbox{\rm m} \; ) \la . . . . . . AB} = - {1\over4} \oint_{\sf ABCD} \Vec{B} \cdot d\Vec{s} = {1\over4} \mu_0 I$, where the - sign indicates that the direction of $\Vec{B}$ is generally from B to A (rather than from A to B) by the right-hand rule. All this gives $\langle B_\parallel \rangle_{\sf AB} = {1\over4} \mu_0 I / 3= -{1\over4} 4\pi \times 10^{-7} \times 2001/ 3$ or .   $(b)$ The field at the midpoint of AB points toward A (by symmetry and the right-hand rule) and has a magnitude given by the usual formula for the field due to a long straight wire, namely .