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Discussion Question 8A P212, Week 8 Magnetic Forces and Torques on Loops A triangular loop with hypotenuse a and 30˚ - 60˚ - 90˚ angles lies in the xy plane. A constant current I circulates clockwise around the loop, and a uniform magnetic field of strength B points in the +x direction. a = 3.0 cm B = 0.04 T I=5A y B (a) Calculate the forces on each of the three sides of the current loop (AB, BC, and CA), and sum them to find the net force on the loop. Remember that you will be adding vectors, so express each force in terms of its components. C 60° a I B 30° A x Since dF=i d B , the force will to both d and B which means it will lie along or against the Z axis for each current segment. By the right-hand rule the force on AB segment will point into the plane of the paper or in the -zˆ direction ; while the force on the AC segment will point into the plane of the paper or in the +zˆ direction. Since B is AB we have FAB =Bi AB =(0.04 T)(5 A) (0.03 m cos 300 ) FAB 0.0052 N zˆ Since segment BC is || to B its force vanishes because of cross prod FBC 0 Segment CA is at 600 wrt B thus FCA =Bi CA sin600 =Bia sin 600 FCA 0.0052 N zˆ (b) Inspect the directions you found for the forces on each of the three sides. These forces are clearly trying to turn the loop … in which direction does the resulting torque point? The left side is being pushed into the plane of the paper while the hypotenuse is being pushed out of the plane By the RH rule, the torque is about -yˆ (c) What is the magnetic moment of the loop? Since is a vector, remember to specify its direction. i area.The area of the loop is area=1/2 (base) (height). Call the height AB a cos 600 =a/2 and the base BC a sin 600 =a 3/2. 2 1 a a 3 ia2 3 5A (.03m ) 3 Then i = = =9.74e-4 A m2 By RH rule 8 2 2 2 8 this is in -zˆ direction. Thus 0.000974 A m2 zˆ (d) What is the torque on the loop? Since is a vector, remember to specify its direction. Is it the same as what you found in part (b)? I hope so … and I hope you’ll agree that using magnetic moments to analyze the loop is a lot easier than working with forces directly! B 0.000974 (.04) zˆ xˆ 3.90e 5 N zˆ xˆ 0.0000390 N m yˆ (e) In which direction would you orient the B field so that no torque was exerted on the loop? Since B , we want || B to make =0 B || zˆ (f) What is the potential energy U of the loop? Is U at its maximum or minimum possible value, or is it somewhere in between? U B B ( zˆ xˆ ) 0 ; Umax =+ B and Umin - B hence U is in between Now imagine that the AB side of the loop is fixed to a pivot, and that the loop is tilted at an angle = 35˚ with respect to the xy plane. (Picture point C raised out of the paper to make the specified angle). (g) Does the tilt change the magnetic moment of the loop? If so, specify the new vector . z . AB x C x 350 μ 550 is unchanged but now it picks up an x and z component. From the diagram cos 550 xˆ sin 350 zˆ 0.000974 cos 550 xˆ sin 350 zˆ 0.000559 xˆ 0.000798zˆ (h) What is the torque on the loop in this orientation? 2 sin 550 and points out of paper plane or -yˆ direction 0.000974 (.04) sin 550 yˆ 3.19 10-5 N m yˆ (i) What is the potential energy U of the loop in this orientation? U U B B cos 550 2.235 105 J +900 (j) What tilt angle would produce zero torque on the loop? It’s important to note that there are two such equilibrium angles. One produces stable equilibrium, the other produces unstable equilibrium. Which is which? -900 Generalizing part (h) sin 900 cos Thus we are looking for zeros of cos which are 900 To analyze stability we need to plot U( . U - - cos 900 or U - sin As plot shows, the minimum (stable equilibrium) is at (k) Finally, suppose the triangular loop had N turns of wire instead of just one. How would this change the quantities , , and U? multiply by N for each 3 Discussion Question 8B P212, Week 8 The Magnetic Field due to Current Loops and Infinite Wires In lecture 14, the B field of an infinite straight wire was 0 I 2 107 I presented: B . The derivation of this formula 2 r r from the Biot-Savart law is given in Appendix A of the lecture notes. We won’t go through it here, since you will soon learn a much easier way to obtain the same result (Ampere’s Law!). Instead, let’s just use the formula, and our knowledge of the direction of the magnetic field around a wire. y B A a x a Four long parallel wires are located at the corners of a square of side a. Each wire carries a current I. The top two currents are directed into the page, and the bottom two out of the page. a = 15 cm I = 2.5 A (a) Calculate the force per unit length exerted on the wire in the top-right corner by the field of the other three. Remember to give both the magnitude and direction of this force. One approach is to find the magnetic forces on the top-right corner due to the other three corners and add these three forces by components. . #3 y F1 #1 #2 y #1 #1 B2 a a a a x #3 #2 a x The diagram on the left shows the forces and fields on the top-right corner due to corner #1 BIL The force in this case is along the direction between the wires. The force/length will be F L 7 2 F1 I 2 10 I F 2 107 8.333 / m . Similarly F2 8.333 / m or IB I r L a L 2 I 8.333/ 2 . Diagram on right shows the direction of these 3 forces. and F3 2 107 a 2 F1 F2 F3 8.333 xˆ yˆ 8.333 / m 8.333 / m xˆ ; 8.333 / m yˆ and Hence xˆ yˆ L L L 2 2 2 2 F F 8.333(1 1/ 2) xˆ 8.333(1 1/ 2) yˆ -4.17 xˆ 12.5 yˆ / m Thus L L y (b) Calculate the magnetic-field vector B at the center of the square (point A). The key is superposition, which works for both E and B fields. #4 #1 #2 #3 a a #4 #1 #2 #3 x a 2 I 2 I Diagram shows direction of all four fields. Each field B r a 2/2 B 2 I The yˆ components cancel ; the four xˆ components add. Each Bx a 2 2 107 I 8 107 (2.5 A) xˆ Bx 4 13.33 a (0.15m) z A circular wire loop of radius R carries a current I circulating in the counterclockwise direction. The loop lies in the xy plane, and is centered on the origin. We are interested in the magnetic field at a point A located on the z axis a distance d above the xy plane. A d y R (c) What is the magnetic field B at the point A? (i) x I The only way to solve this is with the Biot-Savart law, in 2 all its integral glory. We will have to integrate all the contributions dB to the field from each little current segment dI in the wire loop. As our first step, what is the direction of the total B field at point A? B The diagram on the right shows the direction of the magnetic field due to the wire segment that is passing through the x-z plane. As you revolve this current segment around the xy plane, only the z component survives. z 90- d 2 R2 d x R (ii) From step (i), we realized that only one component will contribute to the final magnetic field. Good ... now write down an expression for this component of dB, using the Biot-Savart law. d I d rˆ I d I 0 ˆ from the diagram we see sin 90 = cos zˆ dB dB z 4 r 2 4 r 2 4 r 2 I d I R d R R ˆ dB r d 2 R 2 ; cos z zˆ 2 3 d 2 R2 d 2 R2 4 d 2 R 2 4 d 2 R 2 (ii) Finally, do the integral. To accomplish this, rewrite your integrand dB so that it depends on only one integration variable. d is a path along circumference thus d =Rd as defined in figure. B dB B 4 2 I R 2 4 d R 2 2 I R zˆ d 2 R2 3 zˆ 2 3 2 0 4 I R 2 d R 2 2 3 I R 2 d 2 R2 3 zˆ d zˆ 3 Discussion 8C P212, Week 8 Small babies, huge magnets, and dangerous currents A current loop that is 1m long and 0.5 m wide carries a current of 100 A and has 50 turns. It is pivoted to turn frictionlessly on an axle along the y-axis. The second pivot support is not drawn. The coil is bathed in a magnetic field of 1.2 T that points either parallel or anti-parallel to the x-axis (you decide from the figure). The moment of inertia of the coil about this pivot is I=2 kgm2. The baby-eye view of the coil orientation is shown to the left of the baby. Initially meaning the coil is in the x-y plane. 1m z 0.5m y N B=1.2 T n=50 turn i=100 A I=2.0 kg•m2 S x a) Find the initial forces on each of the 4 sides of the coil when it lies in the x-y plane as shown. Specify their magnitude and direction. Only coil segments with currents along the +yˆ and -yˆ will have forces. The magnitudes will be F nBiL 50 1.2 100 A 0.5m 3000 N . The force on the right segment (near the south pole) ˆ will be directed up along +zˆ . The force on the left segment (near north pole) will along -z. b) Find the direction and magnitude of the magnetic moment when the coil is in the x-y plane in terms of components. By right hand rule points along the -zˆ axis with strength 2500 zˆ m 2 ni area 50 100 A 1m 0.5m c) Find the torque on the coil when it is in the x-y plane. Specify the direction of the torque. Is it consistent with your answers to part a)? xˆ yˆ zˆ 0 0 1 3000 yˆ Nm. It turns along -y axis which causes to increase 1 0 0 From r F we have =2 3000N 0.5m 3000 N m which agrees. d) The baby tries to hold the coil in the x-y plane and prevent it from turning. Can he do it? Can his father do it? Can King-Kong do it? Calculate the force required to hold the coil in the x-y plane if the force is applied on the coil side that is closest to the South pole of the magnet. To counter the magnetic torque a force of Fr must be applied. By the rules of the problem r 0.5 m and Nm F=3000/0.5 = 6000 N. This would need to applied in the -zˆ direction The force is equivalent to a baby lifting 6000/9.8 = 612 kg. Clearly much more than the baby or father can lift or hold down. Call in the big gorilla. Find the components of when the coil is inclined at an angle When 0, is in the -zˆ direction. As increases rotates in the +xˆ direction. This means sin xˆ cos zˆ 2500 sin xˆ cos zˆ Am 2 e) King-Kong releases the coil (from rest) and it begins to turn. Does increase or decrease? d as a function of . What range does the coil Describe the motion in terms of dt motion cover? At what is the largest? Use the convention that 0 . Recall that 1 the kinetic energy of a rotating object is KE I . U 2 900 1800 U B 1.2 xˆ 2500 sin xˆ cos zˆ 3000sin J. The diagram shows U as a function of Since coil is released from rest with U = 0, it oscillates between 0 and 1800 . The maximum KE or occurs at bottom of well or at 900. f) Find the maximum value of that the coil achieves. Use conservation of energy E 0 KE U 0 E90 1 I 3000 2 2 3000 J 2 kg m 2 54.8 s 1 g) Find the maximum speed of the coil segment that lies along the + y axis after the coil is released. v r m 54.8 s 1 27.4 m/s or about 56 miles per hour. Baby beware!! 2 Discussion 8D P212, Week 8 Magnetic Dipoles in Atomic Physics Here is one reason that we are studying magnetic dipole moments: nature is filled with current loops at the atomic and subatomic levels! Magnetic dipoles appear constantly in those fields of physics. Consider the hydrogen atom (one proton (charge +e) with an electron (charge –e) orbiting around it). A proper description of the hydrogen atom requires quantum mechanics (Physics 214!), but you can gain a feeling for its behavior using what you already know from Physics 211 sz = +1/2 and 212. In this problem, you will study its magnetic properties, It is a reasonable approximation to consider the electron to be in a circular orbit of radius rB (the “Bohr radius) = 5.3 ×10−11 m. The charge of the electron is −e =−1.60 ×10−19 C and its mass is = me 9.11×10−31 kg . The electron charge of the proton is +e , and is mass ( 1.67 ×10−27 kg) is so large that we may consider it to be a fixed center of the electron’s orbit (like the Earth-Sun system). lz = +2 (a) What is the orbital speed of the electrons? 2 = = F KQ me v 2 / rB . 1Q2 / rB Hence, v 2 KQ1Q2 / me rB = 9 ×109 × (1.6 ×10−19 ) 2 = 0.4772 ×1013 m 2 / s 2 −31 −11 (9.11×10 )(5.3 ×10 ) = v 2.184 ×106 m / s. (b) What is the angular momentum of the electron? | L | =| r × p | = mvr = (9.11×10−31 ) × (2.184 ×106 ) × (5.3 ×10−11 ) =1.0545 ×10−34 J s (Note that = 1.054 ×10−34 J s ) (c) Think of electron’s orbit as a current loop, what is the magnetic moment of this loop? m =IA =(e v evr )(π r 2 ) = =(1.60 × 10−19 ) × (2.184 × 106 ) × (5.3 × 10−11 ) / 2 =9.26 × 10−24 A m 2 2π r 2 (d) The electron can be in other orbits as well. Show that the magnetic moment of each circular orbit is proportional to the angular momentum. The constant of proportionality is called the “gyromagnetic ratio”. You will find that it depends only on the mass and charge of the electron, so it is a property of the electron (i.e., valid in all situations). / 2 (q / 2m)(= is g q / 2m m qrv mrv) ( q / 2m) L; Hence, the g -factor= = = (e) Now, put the atom in a uniform field, B = 1 T. Calculate the difference between the magnetic energies when the field is pointing up and down (perpendicular to the plane of the elecron’s orbit). ∆U= 2 µ B= 1.852 ×10−23 J z