familiari (nf3795) – Magnetic Force and Field – yeazell – (58395) 1This print-out should have 11 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsConsider two parallel wires where the mag-nitude of the left current is 2 I0and that ofthe right current is I0. Point A is midway be-tween the wires, and B is an equal distanceon the other side of the wires.A•B•The ratio of the magnitude o f the magneticfield at point A to that at point B is1.BABB= 02.BABB=233.BABB=434.BABB= 35.BABB=526.BABB= 27.BABB= 48.BABB=139.BABB=1210.BABB= 9 correctExplanation:Basic Concepts: Magnetic Field due to aLong Wire:B =µ0I2 π r.Principle of sup erposition.Right-hand rule.Solution: Let’s call the distance betweenthe wires r. The ma gnetic field at A due tothe upward current isBup,A=µ0(2 I0)2 π (r/2)=2 µ0I0π r.The right-hand rule tells us the direction isinto the paper. Due to the fact that A isthe same distance from both wires, the totalmagnetic field at A isBA=2 µ0I0π r+µ0I0π r=3µ0I0π r.Now, the field at B due to the upward currentisBup,B=2µ0I02 π (3 r/2)=2µ0I03 π ragain into the paper, while the downwardcurrent givesBdown,B=µ0I02 π (r/2)=µ0I0π rout of the paper. So at B, the net field out ofthe paper is:BB= Bdown,B− Bup,B=µ0I0π r1 −23=µ0I03 π r.Comparing their magnitudes, we findBABB=3 µ0I0π rµ0I3 π r=9 .002 10.0 pointsConsider a long wire and a rectangular currentloop.familiari (nf3795) – Magnetic Force and Field – yeazell – (58395) 2ABCDI1ℓbaI2Determine the magnitude and direction ofthe net magnetic force exerted on the rectan-gular current loop due to the current I1in thelong straight wir e above the loop.1.~F =µ0I1I2ℓ2 π(a −b), right2.~F =µ0I1I22 π(a −b),left3.~F =µ0I1I2ℓ2 πaa + b, down4.~F =µ0I1I2a2 π ℓ(b −a), up5.~F =µ0I1I2ℓ2 π(a + b), up6.~F =µ0I1I22 πa ba + b, down7.~F =µ0I1I2ℓ2 πa b, down8.~F =µ0I1I2ℓ2 πba (a + b), up correctExplanation:To compute the net force on the loop, weneed to consider the forces on segments AB,BC, CD, and DA. The net force o n the loopis the vector sum of the forces on the pieces ofthe loop. The magnetic force onAB due tothe straight wire can be calculated by using~FAB= I2ZBAd~s ×~B .In order to use this, we need to know themagnitude and direction o f the magneti c fieldat each point on the wire loop. We can applythe Biot-Savart Law. The result of this is thatthe magnitude of the magnetic field due to thestraight wire isB =µ0I12 π r,and the direction of the mag netic field is givenby the right hand rul e; the field curls aroundthe straight wire with the field coming out ofthe page above the wire and the field goinginto the page below the wire. We can nowfind the force on the segmentAB; applyingthe right hand r ul e to find the direction of thecross product, d~s×~B, we see that the force willbe in the up direction. Since the wire alongthe segmentAB is straight and always at aright angl e to~B, the cross product simplifiesto B ds. Since the magnitude of the magneticfield is constant al ong segmentAB, it cancome out of the integra l which simplifies togive us the result,FAB= I2ℓ B1= I2ℓµ0I12 π a.Following t he same argument, we see that theforce on the segment CD isFCD= I2ℓµ0I12 π (a + b),and it s direction is down. This is becausethe directio n of the current is now in in theopposite direction along segmentCD!We can do the use the same procedure forsegmentsBC and DA, but because the mag-netic field decreases wi th distance from thestraight wire,~B is changing along these seg-ments. This means that the integrals are notas simple. Using the right hand rule, we seethat the force on segmentBC is directed to-wards the right and the force on segment DAis directed towards the left. Because the twosegments of wire are symmetrically placed,their magni tudes will be equal. Since theseforces on the loop have equal magnitudes butopposite directions, they will cancel. Look-ing more cl osely at segmentsBC and DA,we see that for each small portion of the seg-mentBC, we can find a smal l portion on thesegment DA such that the forces on thesetwo portions are the same magnitude (Thesmall po r tions a re the same distance from thefamiliari (nf3795) – Magnetic Force and Field – yeazell – (58395) 3straight wir e. ); because the current s are in op-posite directio ns, the forces on them w ill bein opposite directions. Their contributions tothe net force cancel.The net force on the loop is then~Floop=~FAB−~FCD= I2ℓµ0I12π a− I2ℓµ0I12 π (a + b)=µ0I1I2ℓ2 πba (a + b).A BCDI1I2I2I2I2I2ℓ × B1FAB⊗I2ℓ × B1FCD⊗Here, a positive force is in the up direction,and a negative force is in the down direction.So, the direction of the net force is up.003 (part 1 of 2) 10.0 pointsA circular current loop of radius R is placedin a horizontal plane and maintains a currentI. There is a constant magnetic field~B inthe xy-plane, with the angle α (0◦< α < 90◦)defined with respect to y-axis. The current inthe loop flow s clockwise as seen from above.Izˆkxˆıy, ˆBαWhat is the direction of the torque vector~τ exerted on the current loop by the field?1. bτ =ˆk −ˆ sin α2. bτ =ˆk −ˆı√23. bτ = −ˆk4. bτ = +ˆ5. bτ = +ˆı6. bτ =ˆı +ˆk√27. bτ = −ˆı8. bτ = +ˆk correct9. bτ = −ˆ10. bτ = ˆı + ˆ sin αExplanation:µBIzˆkxˆıy, ˆαˆ ׈ = 0 and −ˆ ׈ı =ˆk, so the torq ue is~τ = ~µ ×~B = µ (−ˆ) × [Bx(+ˆı) + By(−ˆ)]= µ Bxˆk ,with direction +ˆk . This agrees wit h the an-swer from the right-hand rule.004 (part 2 of 2) 10.0 pointsWhat is the magnitude of the torque vector?1. k~τ k = I π R2B sin α correct2. k~τ k = I π R2B sinπ2− α3. k~τ k = I R2B sin α4. k~τ | = I π R2B sinπ25. k~τ k = I 2 π R2B sinπ2− α6. k~τ k = I π R2B sinπ2+ α7. k~τ k = I 2 π R2B sin αfamiliari (nf3795) – Magnetic Force and Field – yeazell – (58395) 48. k~τk = I R2B sinπ2+ α9. k~τk = I R2B sinπ2− α10. k~τk = I 2 π R2B sinπ2+ αExplanation:The magnetic dipole moment isµ = I A = I π R2and the magnitude of a cross product isk~Ck = k~A ×~Bk = A B sin θ ,where θ is the angle between~A and~B .The angle between ~µ and~B is α , so themagnitude of the torque isτ = µ B sin θ = I π R2B sin α .005 10.0 pointsA cross