Apr 2008

Physics Challenge for Teachers and Students Solution to April 2008 Challenge ◗ A BLT Sandwich Challenge: A square loop made of wire with negligible resistance is placed on a horizontal frictionless table as shown (top view). The mass of the loop is m and the length of each side is b. A nonuniform vertical magnetic field exists in the region; its magnitude is given by the formula B = B0 (1 + kx), where B0 and k are known constants. The loop is given a quick push with an initial velocity v along the x-axis as shown. The loop stops after a time interval t. Find the inductance L of the loop.

b b

using Ohm’s law in the last step.1 But we are told that the resistance R of the loop is zero. Therefore b2dB / dt = LdI / dt. (That is, the back emf exactly cancels the externally induced emf.) Integrating both sides with respect to time then implies that b2B = LI + c, where c is some constant that depends on the initial conditions. Let’s assume that the loop starts out with zero current before it is placed on the table into the magnetic field. Then c = 0 and the net flux, ext – back, linking the loop is always zero. We conclude that the current in the loop is

I=

v

b2 B0 (1 + kx ) L

(2)

in the clockwise direction. But we now have four wire segments (i.e., the four sides of the Solution: Suppose that the external magnetic field loop) carrying current I in a magnetic field B(x). Bext  B = B0(1 + kx) is directed out of the page. Consequently there is a magnetic force on each As the loop moves to the right, the externally of them. The forces on the top and bottom sides 2 linked flux through the loop ext = Bextb increas- are equal and opposite. That on the right side is es; its time derivative results in an emf of –IbB(x), where the minus sign indicates that it is ext = –d ext /dt, where the minus sign indicates directed leftward since I is down the page and B that it drives a current I around the loop in the is out of the page, and where we let x measure the clockwise direction. Noting that dBext /dt = B0k position of the right side of the loop. Thus the left dx/dt is not constant (since the speed dx/dt of the side is located at position x – b and the force on loop decreases monotonically from υ initially to it is +IbB(x – b). Consequently the net magnetic 0 finally), the current I cannot be constant either. “braking” force on the loop is We therefore have a time-varying clockwise curF = −Ib {B0[1 + k x ]− B0[1 + k( x − b )]} rent in an inductor L and so we generate a back emf back = –LdI / dt  –d back/dt in the coun(3) 4 2 kb B 0 ter-clockwise direction. Consequently the net =− (1 + kx ) L clockwise emf around the loop is after substituting Eq. (2). dI 2 dB (1) ε =b − L = IR dt dt We can write this result in the form F = –K (x – x0), where K = k2b4B20/L and x0 = –1/k x

THE PHYSICS TEACHER ◆ Vol. 46, 2008

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are constants. This has the form of a Hookean restoring force, where the effective spring constant is K and the equilibrium position is located at x0. Assume that the loop is originally placed at the equilibrium position so that it doesn’t move before it is given the impulsive push. The loop will subsequently oscillate about that position forever after it is pushed, since there is no resistive dissipation of energy. It first reaches the rightward turning point in a time t equal to one-quarter of a period T = 2π m / K so that 2

mL π t= 2 2 k b 4 B02

⇒

1 ⎛ 2kb 2 B0t ⎞⎟⎟ L = ⎜⎜⎜ ⎟ . m ⎜⎝ π ⎟⎠ (4)

By using the standard textbook expression for the magnetic field of a straight wire segment, one can fairly easily show that the inductance of a square loop is 1 1

L=

2μ0b π

∫∫

r /b 0

xdxdy 2

y x +y

2

,

(5)

where r is the radius of the wires and is assumed to be small compared to b. This integral works out to be approximately

L≈

2μ0b ⎛⎜ b ⎞⎟ ln ⎜⎜ ⎟⎟ , ⎝r ⎠ π

(6)

which diverges if we let r go to zero. Instead, taking r = b/1000, this result gives L  4bμ0, where μ0 = 0.4π μH/m is the permeability of free space. For example, if b = 1 m then L  5 μH. The important point is that Eq. (4) then implies that

t2 ∝

m b(ΔB )2

,

(7)

assuming r << b, where ΔB  B0kb is the variation in the external magnetic field strength across the width of the loop. The stopping time increases if the loop’s mass (inertia) is increased, while the time decreases if the size of the loop or the magnetic field gradient is made larger.

Addendum I would like to thank José Íñiguez for pointing out that calculating the externally linked flux as ext = Bextb 2 is not correct because the external magnetic

THE PHYSICS TEACHER ◆ Vol. 46, 2008

field varies across the length of the loop. Instead, the external flux is

Φext = ∫ B dA =

x +b

∫

Bext ( x ′ )bdx ′

(8)

x

whose time derivative is

d Φext dx d = dt dt dx =υ

x +b

∫

Bext ( x ′ )bdx ′

(9)

x

ΔB 2 b . Δx

But because B is a linear function of x, ΔB/Δx = dB/dx = B0k, , and so Eq. (1) and the remainder of the solution is nevertheless correct. (Contributed by Carl E. Mungan, U. S. Naval Academy, Annapolis, MD) 1.

In traditional textbook presentations of Faraday’s law for the current induced in a loop placed in a varying magnetic field, the loop’s inductance is assumed to be negligible so that ext = IR. In the present problem, the loop’s resistance is instead negligible. More generally, Eq. (1) can be viewed as the equation of an LR series circuit with ext playing the role of the usual battery.

We would also like to recognize the following contributors: Sanjeet Singh Adarsh (PTC, Pune, India) Marianne Breinig (The University of Tennessee, Knoxville, TN) Alan J. DeWeerd (University of Redlands, Redlands, CA) F. Javier Doblas (Escuela Técnica Superior de Ingenieros, Sevilla, Spain) Don Easton (Lacombe, Alberta, Canada) Bruce Gordon (Kimball Union Academy, Meriden, NH) Fredrick P. Gram (Cuyahoga Community College, Cleveland, OH) Art Hovey (Milford, CT)

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J. Iñiguez (Universidad de Salamanca, Salamanca, Spain) John Mallinckrodt (Cal Poly Pomona, Pomona, CA) Stephen McAndrew (Trinity Grammar School, Summer Hill, NSW, Australia) Bayani I. Ramirez (San Jacinto College South, Houston, TX) Finally, we would like to acknowledge Bartley L. Cardon (MIT Lincoln Laboratory, South Portland, ME), whose name was omitted from the list of the March Challenge solvers, and apologize to Dario Castello (Vasquez High School, Acton, CA), whose last name was misspelled there. Many thanks to all contributors and we hope to hear from you in the future! Please send correspondence to: Boris Korsunsky [email protected]

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THE PHYSICS TEACHER ◆ Vol. 46, 2008