KORUNSKY question
2
Physics Challenge for Teachers and Students ◗ A Faradayan Slip 1 Two identical conducting bars rest on two horizontal parallel conducting rails. The bars are perpendicular to the rails and parallel to each other as shown. The distance between the bars is l. At a certain moment, a uniform vertical upward magnetic field is turned on. The field quickly reaches its maximum magnitude and then remains constant. Neglecting friction, find the new distance between the bars. Assume that the resistance of each bar is much greater than the resistance of the rails. 1. Adapted from Physics Olympiads, by A.I. Slobodyanuk, L.G. Markovich, and A.V. Lavrinenko, published by Aversev, Minsk (2003) (in Russian). Solution: First, it should be noted immedi- ately that a trivial dimensional analysis reveals that the answer can only be zero or a numerical multiple of l. It’s not hard to guess (see my post- script) that the answer might well be l 2 . The somewhat harder part is proving it. The changing flux due to the magnetic field B out of the page induces an emf that drives a current I clockwise (by Lenz’s law) around the loop. As a result, magnetic forces F act on the rods, causing them to move closer together. Let- ting the separation of the rods at any specific time be x, we find by symmetry that the accel- eration of each rod is –x .. /2. Thus, by Newton’s second law, the expression for the force on a segment of wire, Ohm’s law, and Faraday’s law Solutions to April 2006 Challenge (respectively), we have x F m m IwB wB m R =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ( ) = − ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎛ ⎝ 2 2 2 2 ε ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ( ) =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ + wB mR d dt xwB wB mR xB 2 ( xB ), where m and R are the mass and resistance of each rod and w is the separation of the rails. Because the field is turned on “quickly,” we may assume that it reaches its maximum value while x . remains small and before x has a chance to change very much from its initial value, l. Thus, during the “turning on” phase the second term in Eq. (1) is x wB mR lB wl mR d dt B ≈− ( ) =− 2 2 2 2 , which can be integrated over the brief time that the field is changing to find x V w lB mR final final ≡− =− 2 2 2 . At later times the first term in Eq. (1) is strictly zero so that x wB mR x =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ 2 2 final , showing that the rod subsequently decelerates to rest. This relationship is trivially integrated once (starting at the beginning of the constant B phase) to find that x V wB mR x x V x l −− ( ) =− ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ ⇒ = − 2 2 2 1 final . From which x final (the value of x when x . = 0) l (top view) 94 THE PHYSICS TEACHER ◆ Vol. 44, 2006
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Transcript of KORUNSKY question
Physics Challenge for Teachers and Students
◗ A Faradayan Slip1 Two identical conducting bars rest on two horizontal parallel conducting rails. The bars are perpendicular to the rails and parallel to each other as shown. The distance between the bars is l. At a certain moment, a uniform vertical upward magnetic field is turned on. The field quickly reaches its maximum magnitude and then remains constant.
Neglecting friction, find the new distance between the bars. Assume that the resistance of each bar is much greater than the resistance of the rails.
1. Adapted from Physics Olympiads, by A.I. Slobodyanuk, L.G. Markovich, and A.V. Lavrinenko, published by Aversev, Minsk (2003) (in Russian).
Solution: First, it should be noted immedi-ately that a trivial dimensional analysis reveals that the answer can only be zero or a numerical multiple of l. It’s not hard to guess (see my post-script) that the answer might well be l
2. The
somewhat harder part is proving it.
The changing flux due to the magnetic field B� out of the page induces an emf � that drives a current I clockwise (by Lenz’s law) around the loop. As a result, magnetic forces F act on the rods, causing them to move closer together. Let-ting the separation of the rods at any specific time be x, we find by symmetry that the accel-eration of each rod is –x../2. Thus, by Newton’s second law, the expression for the force on a segment of wire, Ohm’s law, and Faraday’s law
Solutions to April 2006 Challenge
(respectively), we have
��xFm m
IwB
wBm R
= −⎛⎝⎜⎜⎜
⎞⎠⎟⎟⎟ = −
⎛⎝⎜⎜⎜
⎞⎠⎟⎟⎟( ) =
−⎛⎝⎜⎜⎜
⎞⎠⎟⎟⎟⎛⎝
22
2
2
ε⎜⎜⎜⎜⎞⎠⎟⎟⎟ = −
⎛⎝⎜⎜⎜
⎞⎠⎟⎟⎟ ( )
= −⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟ +
wBmR
ddt
xwB
w BmR
xB2
( � �xxB),
where m and R are the mass and resistance of each rod and w is the separation of the rails.Because the field is turned on “quickly,” we may assume that it reaches its maximum value while x. remains small and before x has a chance to change very much from its initial value, l. Thus, during the “turning on” phase the second term in Eq. (1) is
�� �xw BmR
lBw lmR
ddt
B≈ − ( )= −2 2
2
2,
which can be integrated over the brief time that the field is changing to find
�x Vw lB
mRfinalfinal≡ − = −
2 2
2.
At later times the first term in Eq. (1) is strictly zero so that
�� �xw B
mRx= −
⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟
2 2
final ,
showing that the rod subsequently decelerates to rest.
This relationship is trivially integrated once (starting at the beginning of the constant B phase) to find that
��
x Vw B
mRx
xV
xl
− −( ) = −⎛
⎝⎜⎜⎜⎜
⎞
⎠⎟⎟⎟⎟⎟
⇒ = −2 2
21
final
.
From which xfinal (the value of x when x. = 0)
l
(top view)
94 THE PHYSICS TEACHER ◆ Vol. 44, 2006
can immediately be read off.
xl
final =2
Postscript: As mentioned at the start, this result might well have been anticipated from a combination of 1) the dimensional analysis argument, 2) the fact that the rods clearly get closer together, and 3) an analogy with other problems in which reaching a new equilibrium involves energy dissipation.
For instance, when a fully charged capacitor is connected to an identical uncharged capaci-tor, half the energy is lost and the final potential difference is half the original value. This result depends on there being resistance in the system. Without it we get LC oscillations with each capacitor’s potential difference oscillating about half the original value.
Similarly, in this problem the resistance is vital in order to allow the magnetic flux through the loop to change from its original value, zero. If the rods had moved to conserve the flux, they would have ended up at a separation of zero, with the resistance they moved just half that far, and ended up with half the flux they would have had if they hadn’t moved at all. The answer doesn’t depend on R as long as there is some R.
(Contributed by A. John Mallinckrodt, Cal Poly Pomona, Pomona, CA)
THE PHYSICS TEACHER ◆ Vol. 44, 2006 95
Many other readers also sent us the solutions, although not all were correct. We would like to recognize the following contributors:
Marianne Breinig (The University of Tennessee, Knoxville, TN)
Don Easton (Lacombe, Alberta, Canada)Fredrick P. Gram (Cuyahoga Community
College, Cleveland, OH)Art Hovey (Milford, CT)David Jones (Miami Palmetto Sr. High School,
Miami, FL)J. Iñiguez (Universidad de Salamanca,
Salamanca, Spain)Stephen McAndrew (Trinity Grammar School,
Summer Hill, NSW, Australia)Carl E. Mungan (U.S. Naval Academy,
Annapolis, MD)
Many thanks to all contributors and we hope to hear from you in the future!Please send correspondence to: Boris Korsunsky 444 Wellesley St. Weston, MA 02493-2631 [email protected]
