Magnetic Dipoles Challenge Problem Solutions

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Christine Cole
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1 Magnetic Dipoles Challenge Poblem Solutions Poblem 1: Cicle the coect answe. Conside a tiangula loop of wie with sides a and b. The loop caies a cuent I in the diection shown, and is placed in a unifom magnetic field that has magnitude B and points in the same diection as the cuent in side OM of the loop. At the moment shown in the figue the toque on the cuent loop a) points in the î -diection and has magnitude IabB /. b) points in the + î -diection and has magnitude IabB /. c) points in the ĵ -diection and has magnitude IabB /. d) points in the + ĵ-diection and has magnitude IabB /. e) points in the î -diection and has magnitude IabB. f) points in the + î -diection and has magnitude IabB. g) points in the ĵ -diection and has magnitude IabB. h) points in the + ĵ-diection and has magnitude IabB. i) None of the above. Poblem 1 Solution: b. The magnetic dipole moment vecto is μ = Iab /ĵ. The toque on the cuent loop is then τ = μ B = (Iab / ) ˆj Bk ˆ = (IabB / ) i ˆ.

2 Poblem : A wie ing lying in the xy-plane with its cente at the oigin caies a counteclockwise cuent I. Thee is a unifom magnetic field B = Bî in the +x-diection. The magnetic moment vecto μ is pependicula to the plane of the loop and has magnitude μ = IA and the diection is given by ight-hand-ule with espect to the diection of the cuent. What is the toque on the loop? Poblem Solution: The toque on a cuent loop in a unifom field is given by τ = μ whee μ =IA and the vecto μ is pependicula to the plane of the loop and ight-handed with espect to the diection of cuent flow. The magnetic dipole moment is given by B, μ = IA = I(π R kö) = π IR kö. Theefoe, τ = μ B = ( πir kö ) (B ö i)= πir B ö j. Instead of using the above fomula, we can calculate the toque diectly as follows. Choose a small section of the loop of length ds = Rdθ. Then the vecto descibing the cuent-caying element is given by The foce df that acts on this cuent element is Id s = IRdθ ( sin θ î +cos θ ĵ) d F = Id s B = I Rdθ ( sin θ ˆ i + cos θ ĵ ) ( Bî) = IRB cosθ d θkˆ The foce acting on the loop can be found by integating the above expession. F = df = π ( IRBcosθ)dθ k ö = IRB sinθ π k ö = We expect this because the magnetic field is unifom and the foce ona cuent loop in a unifom magnetic field is zeo. Theefoe we can choose any point to calculate the toque about. Let be

3 the vecto fom the cente of the loop to the element Id s. That is, = R (cosθî + sin θ ĵ). The toque dτ = df acting on the cuent element is then dτ = df = R (cosθ ö i + sinθ j ) ( IRBdθ cosθ k ö ) ( ) = IR Bdθ cosθ sinθ ö i cosθ j Integate dτ ove the loop to find the total toque τ. τ This agees with ou esult above. dτ = = π IR Bdθ cosθ(sinθ ö i cosθ j ) = IR B π (sinθ cosθ ö i cos θ = π IR B j j)dθ

4 Poblem 3: 1. Foce on a Dipole in the Helmholtz Appaatus The magnetic field along the axis of a coil is given by B ( z ) = z N μ I R 1 (z + R ) 3/ whee z is measued fom the cente of the coil. Conside a disk magnet (a dipole) suspended on a sping, which we will use to obseve foces on dipoles due to diffeent magnetic field configuations.

5 (a) Assuming we enegize only the top coil (cuent unning counte-clockwise in the coil, ceating the field quoted above), and assuming that the dipole is always well aligned with the field and on axis, what is the foce on the dipole as a function of position? (HINT: In this situation F z = μ z db z dz ) (b) The disk magnet (togethe with its suppot) has mass m, the sping has sping constant k and the magnet has magnetic moment μ. With the cuent on, we lift the bass od until the disk magnet is sitting a distance z above the top of the coil. Now the cuent is tuned off. Does the disk magnet move up o down? Find the displacement Δz to the new equilibium position of the disk magnet. (c) At what height(s) is the foce on the dipole the lagest? (d) What is the foce whee the field is the lagest? (e) Ou coils have a adius R = 7 cm and N = 168 tuns, and the expeiment is done with I = 1 A in the coil. The sping constant k ~ 1 N/m, and μ ~.5 A m. The mass m ~ 5 g is in the shape of a cylinde ~.5 cm in diamete and ~ 1 cm long. If we place the magnet at the location whee the sping is stetched the futhest when the field is on, at about what height will the magnet sit afte the field is tuned off?. Motion of a Dipole in a Helmholtz Field R R In Pat I of this expeiment we will place the disk magnet (a dipole with moment μ) at the cente of the Helmholtz Appaatus (in Helmholtz mode). We will stat with the disk magnet aligned along the x-axis (pependicula to the cental z-axis of the coils), and then enegize the coils with a cuent of 1 A. Recall that a Helmholtz coil consists of two coils of adius R and N tuns each, sepaated by a distance R, as pictued above. The field fom each coil is given at the beginning of the pevious poblem. (a) The disk magnet will expeience a toque. Will it also expeience a foce? Explain why o why not. Poblem 3 Solutions: 1. Foce on a Dipole in the Helmholtz Appaatus (a)

6 F z =μ z db z / dz That is, towads the coil cente =μ z N μ I R z N μ =μ I R 3 z z (z + R d (z + R ) 3/ dz ) 5/ (b) When the cuent is on, the downwad the gavitational foce and the downwad magnetic foce ae balanced by the sping foce which stetches the sping by an amount Δ l = l i l. When the equilibium has been eached, Newton s Second Law becomes The magnetic foce is F sping + mg k( l i l ) = F N μ ( z = z ) =μ I R 3 z (z + R ) B 5/ Theefoe N μ I R 3z (z + R ) 5/ + mg k( l i l ) = (.1) μ When the cuent is off, and a new equilibium position has been attained, the object moves upwad and the sping is now stetched an amount l f l, theefoe Newton s Second Law becomes mg k( l f l ) =. Solving fo mg = k( l f l ) and substituting into Eq. (.1) yields Thus N μ I R 3z μ (z + R ) 5/ + kl ( f l ) kl ( i l ) = μ N μ I R 3 z (z + R ) 5/ = kl ( i l f )

7 The sping moves upwads a distance l i N μ l I R 3z f 5/ = F mag =μ. (.) k (z + R ) k (c) To find this we just maximize the foce function (find zeos of its deivative): df B d N μ I R 3z dz = dz μ (z + R ) 5/ = z z 5/ 5/ = d ( ( + R ) ) = ( z + R ) 5z z + R ) dz R = z + R 5z z = ± ( 7 / (d) Whee the field is lagest the foce must be zeo. You can eithe think That s whee an aligned dipole would like to be o Maximum field means deivative of field is zeo means no foce. (e) To make the sping stetch the futhest we must be at the location of the lagest foce, a distance z =R/ above the coil (fom c). Fom above the sping will elax by: N μ I R 3z N μ I 3 1 Δ z =μ k (z + R ) 5/ =μ kr ( 1 +1) 5/ 4 7 (.5 A m ) (168)( 4π 1 T m A )(1 A) 33 5/ = 9. mm 1 Nm)(7 cm) 5 ( So the final position is a distance 3.5 cm + 9. mm 4.4 cm above the cente of the coil. Motion of a Dipole in a Helmholtz Field The toque will ceate an angula acceleation: τ =μbsin (θ) = Iα = I θ,

8 which will lead to angula motion. This is a petty ugly diffeential equation to solve, but we can make a stab at it in two diffeent types of appoximations. The fist is to assume a constant toque, pobably not the maximum toque, but maybe half of the maximum. Then we have: We can calculate the field at the cente: B Helmholtz = Δθ = 1 αt = 1 1 μb t. I N μ I R 1 N μ I = 3/ 3/ ((R ) + R ) R ((1 ) +1) ( 168 )( 4π 1 7 T m A -1 )( 1 A ) 8 ( 7 cm ) 5 3/ =. mt = Gauss The moment of inetia of a cylinde is I = mr (5 g )(.5 cm ) = kg m t 4IΔθ 4 ( kg m )( π ) μb 3 (.5 A m )(. 1 T) 9 ms Anothe way of estimating the time is by appoximating the motion as simple hamonic (which it definitely isn t because Δθ is so big). Then the time is a quate of a peiod, which is 8 T π π I π (1.6 1 kg m ) t = = = = 6 ms 3 4 4ω μb (.5 A m )(. 1 T) Note that this should eally be a lowe bound because it is fo small oscillations. Once you get lage oscillations the peiod stats inceasing the eal peiod fo a nonsimple hamonic oscillato with amplitude θ is T π = T SHM 1+ 1!! θ sin + 3!! 4 θ sin + T SHM 1.18!! 4 4!! 4 Meaning the time to otate will be about 7 ms, so ou appoximations wee both petty good. In any case, it will be too fast fo us to see the motion, instead we ll just see the end esult.

9 MIT OpenCouseWae 8.SC Physics II: Electicity and Magnetism Fall 1 Fo infomation about citing these mateials o ou Tems of Use, visit:

Module 18: Outline. Magnetic Dipoles Magnetic Torques

Module 18: Outline. Magnetic Dipoles Magnetic Torques Module 18: Magnetic Dipoles 1 Module 18: Outline Magnetic Dipoles Magnetic Toques 2 IA nˆ I A Magnetic Dipole Moment μ 3 Toque on a Cuent Loop in a Unifom Magnetic Field 4 Poblem: Cuent Loop Place ectangula