FORCE ON A CURRENT IN A MAGNETIC FIELD

Transcription

1 MSN FORCE ON A CURRENT N A MAGNETC FELD FORCE ON A CURRENT N A MAGNETC FELD b Orilla McHarris 1. A Straight Current a. Current Consists of Moving Charges b. Relationship Among Q, v, And c. Force on a Current A Rectangular Current a. Plane of Loop Parallel to b. Loop at an Angle to c. Generaliation to a Current-Carring Coil Acknowledgments A. Geometrical Definition of Vector Product Algebraic Definition of Vector Product C. Direction of the Vector Product Project PHYSNET Phsics ldg. Michigan State Universit East Lansing, M 1

2 D Sheet: MSN Title: Force on a Current in a Magnetic Field Author: Orilla McHarris, Lansing Communit College Version: 3/27/2000 Evaluation: Stage 0 Length: 1 hr; 24 pages nput Skills: 1. Vocabular: current (MSN-0-117); magnetic field, magnetic force (MSN-0-122). 2. Calculate the torque on a rod with respect to a parallel ais, given a constant force perpendicular to the rod (MSN-0-5). 3. Visualie the rotational motion produced b a given torque (MSN-0-33). 4. Calculate the magnetic force on a moving charged particle (MSN ). Output Skills (Knowledge): K1. Derive the epression Q v = l relating a set of charges and their common velocit to their equivalent value as a current. K2. Starting from the Lorent force, derive the epression for the force on a length current-carring wire in a uniform magnetic field. Output Skills (Problem Solving): S1. Calculate the force on a straight section of a current-carring wire in a given uniform magnetic field. S2. Calculate the mechanical torque on a rectangular current loop in a given uniform magnetic field. S3. Calculate the torque on an n-turn rectangular coil in a given uniform magnetic field. Post-Options: 1. The Magnetic Field of a Moving Charge: Magnetic nteractions (MSN-0-124). 2. The Magnetic Field of a Current: The Ampere-Laplace Equation (MSN-0-125). THS S A DEVELOPMENTAL-STAGE PULCATON OF PROJECT PHYSNET The goal of our project is to assist a network of educators and scientists in transferring phsics from one person to another. We support manuscript processing and distribution, along with communication and information sstems. We also work with emploers to identif basic scientific skills as well as phsics topics that are needed in science and technolog. A number of our publications are aimed at assisting users in acquiring such skills. Our publications are designed: (i) to be updated quickl in response to field tests and new scientific developments; (ii) to be used in both classroom and professional settings; (iii) to show the prerequisite dependencies eisting among the various chunks of phsics knowledge and skill, as a guide both to mental organiation and to use of the materials; and (iv) to be adapted quickl to specific user needs ranging from single-skill instruction to complete custom tetbooks. New authors, reviewers and field testers are welcome. PROJECT STAFF Andrew Schnepp Eugene Kales Peter Signell Webmaster Graphics Project Director ADVSORY COMMTTEE D. Alan romle Yale Universit E. Leonard Jossem The Ohio State Universit A. A. Strassenburg S. U. N. Y., Ston rook Views epressed in a module are those of the module author(s) and are not necessaril those of other project participants. c 2001, Peter Signell for Project PHYSNET, Phsics-Astronom ldg., Mich. State Univ., E. Lansing, M 48824; (517) For our liberal use policies see: 3 4

3 MSN FORCE ON A CURRENT N A MAGNETC FELD b Orilla McHarris 1. A Straight Current 1a. Current Consists of Moving Charges. An electric current consists of moving charges. Thus, if a current is placed in a magnetic field, it will be subject to a magnetic force just as single moving charges are. The magnetic force on one charge q moving with velocit v in a magnetic field is: 1 F = q v. (1) The magnetic force on a current is just the sum of such forces on the current s component charges. Assuming these charges all travel with the same velocit: F Total = q 1 v + q 2 v q n v = (q 1 + q q n ) v, = Q v, where Q is the total charge moving with velocit v. 1b. Relationship Among Q, v, And. n order to state the magnetic force in terms of the measured current,, we must find the relationship of Q and v to. The amount of charge dq in a small length of conducting wire dl is given b the overall charge per length, Q/l times the length dl: dq = Q dl (2) l f we hold a current measuring instrument at one point and watch a length dl of charge go b, for all of dl (and therefore, all of dq) to pass the point takes a time dt = dl/v (3) where v is the speed of the charges making up the current. Now the definition of current at a point is the amount of charge per unit time 1 See Force on a Charged Particle in a Magnetic Field (MSN-0-122). MSN passing the point: = dq dt = (Q/l)dl = Qv dl/v l. (4) Remember, however, that the velocit of positive charges and the current are in the same direction, so actuall: l = Q v. (5) For convenience, we will transfer the designation of the current s direction to the length of the straight section of current-carring wire: l = Q v. (6) Remember that the direction of l is alwas to be taken in the direction of the current. 1c. Force on a Current. Now it is eas to state the magnetic force in terms of current. For a set of charges all moving with the same speed in the same direction, F = Q v, (7) or: F = l. (8) Notice that, in a complete electrical circuit, l will have to have several different directions, so different sides of a current loop will in general have different forces on them. The total force on a current loop is then the sum of the force on its separate sides (and if the loop is made up of curved circle, for eample, the total force would be the integral of the small elements of force d F acting on each small length d l). 2. A Rectangular Current 2a. Plane of Loop Parallel to. Let us calculate what will happen to a rectangular loop of current in a uniform magnetic field. First let us take the simple case where the loop is in the - plane and is in the -direction. We must appl Eq. (8) to each of the four sides of the loop separatel (see Fig. 1): a. Side a: Fa = 0 Help: [S-3] b. Side b: Fb = Lẑ; F tends to push side b into the page Help: [S-4] 5 6

4 MSN L d a c W c. Side c: Fc = 0 b ^ ^ ` Figure 1. A rectangular current loop in a magnetic field parallel to the loop s width. d. Side d: Fd = Lẑ; F tends to push side d out of the page. Thus the net result of the four forces is to produce a torque on the current loop about an ais through its center, parallel to the -ais. We can calculate the torque about this ais: 2 Help: [S-6] τ = i τ i = i r i F i = r a F a + r b F b +... (9) = [0 + W 2 L W 2 L]ŷ = W Lŷ = (A ) ŷ where A = W L is the area enclosed b the loop. Suppose the loop starts rotating in response to the torque, resulting in no longer being in the plane of the loop. Then the equations derived above will no longer be valid because the assumed that is in the loop plane. 2b. Loop at an Angle to. Since a torque on a loop will cause it to rotate, we now treat the case where such a rotation has produced an angle θ between and the normal to the plane of the loop (when is in the loop plane, θ = 90 ). 3 Figure 2a shows our rectangular current loop 2 See Force and Torque (MSN-0-5). 3 The normal to the plane of the loop is a unit vector perpendicular to the plane. MSN (a) L d a c W b (b) F on d F on b Figure 2. (a) Oblique view of a rectangular current loop whose normal is at an angle to ; and (b) top view of the current loop. with its normal rotated awa from b some angle θ. Figure 2b shows an overhead view of the same loop. Again we consider the force on each side separatel: a. Side a: Fa = W sin(90 + θ) ŷ = W cos θŷ. The direction of F a is such as to push side a and the entire current loop upward, in the positive -direction. b. Side b: Fb = L ( ẑ). The direction of F b is such to push side b backward, in the negative -direction. c. Side c: F c = W sin(90 θ)( ŷ) = W cos θ( ŷ). The direction of F c is such as to push c and the entire current loop downward, in the negative -direction. d. Side d: Fd = L ẑ. The direction of F d is such as to push d forward, in the positive -direction. Now notice that although the forces on sides a and c are no longer ero, the have equal and opposite effects and are radial; hence the have no effect other than a tendenc to deform the loop if it is not rigid. 4 The forces on sides b and d still operate in such a wa as to produce torques on the loop: τ b = W l sin θ ŷ, 2 4 (1) Adding equal but opposite forces ields ero net force; and (2) an radial force produces ero torque. 7 8

5 MSN and thus the total torque is: τ d = W l sin θ ŷ, 2 τ = W L sin θ ŷ = A sin θ ŷ. t is apparent that the torque on the current loop is a maimum for θ = 90 (i.e. for in the plane of the loop as in Fig. 1) and ero for θ = 0. Thus the tendenc is for a current loop in a magnetic field to rotate until its normal is parallel to, and for it to decelerate as it shoots past that alignment. We could convert this oscillating loop into a rotating one and use it as a means of turning electrical energ into mechanical energ that is, as a DC motor if we could reverse the current in the loop just as its normal becomes parallel to. 2c. Generaliation to a Current-Carring Coil. t should be noted that it is a simple matter to generalie from the force or torque on a single turn loop of current to the force or torque on a man turn coil. n general, coils are wound with the area of each turn the same as all the others and of course the same current would flow through them all. Thus the force or torque on an n-turn coil is generall just n times the force or torque on a one-turn coil. Acknowledgments Kirb Morgan constructed the Problem Supplement. Mark Sullivan gave valuable feedback on an earlier version. Preparation of this module was supported in part b the National Science Foundation, Division of Science Education Development and Research, through Grant #SED to Michigan State Universit. A. Geometrical Definition of Vector Product The vector product of two arbitrar vectors A and is defined as the vector quantit whose magnitude is given b the product of the magnitudes of the two vectors times the sine of the angle between the vectors when the are placed tail-to-tail, and whose direction is perpendicular to the plane formed b A and. The vector product (also referred to as the cross product ) is denoted b A. The magnitude of the product ma be written as: A = A sin θ. MSN There are, however, two directions that are perpendicular to the plane formed b A and. The correct direction ma be chosen b appling the right-hand rule : Rotate vector A into vector through the smaller angle between their directions when the are placed tail-to-tail. Follow this rotation with the curled fingers of our right hand, and the direction of our etended thumb identifies the direction of the vector product. This rule is sufficient to distinguish between the two possible choices for the direction of a vector product. Notice that the order of multiplication in vector products is ver important. The product A has the same magnitude as A, but the directions of the two products are opposite. n general: A = A. We sa that vector products do not commute, or that the vector product is a noncommutative operation. f two vectors are parallel, the angle between their directions is ero, so b the definition of the magnitude of vector products their cross product is ero. Similarl, if two vectors are perpendicular, the angle between their directions is 90. Since sin90 = 1, the magnitude of the vector product of the two is just the product of their magnitudes, and the direction of the vector product is determined b the right-hand rule. appling these observations to the vector product of the cartesian unit vectors ˆ, ŷ and ẑ, we ma derive the following useful relations: ˆ ˆ = ŷ ŷ = ẑ ẑ = 0 ˆ ŷ = ŷ ˆ = ẑ ŷ ẑ = ẑ ŷ = ˆ ẑ ˆ = ˆ ẑ = ŷ. These relations are used in the algebraic definition of vector products.. Algebraic Definition of Vector Product f we epress vectors A and in their cartesian component form, the vector product of A and ma be written: A = (A ˆ + A ŷ + A ẑ) ( ˆ + ŷ + ẑ). f this epression is epanded algebraicall as we would the product ( + 3) (2 5), ecept that the cross product is used instead of scalar 9 10

6 MSN multiplication, then this vector product ma be epressed as a combination of the cartesian components of A and and cross products of the cartesian unit vectors. Using the relations between the cartesian unit vectors developed in Appendi A and denoting the vector product as a third vector C, we ma write: C = A = (A A )ˆ + (A A )ŷ + (A A )ẑ or: C = A A, C = A A, C = A A. The mnemonic for remembering the order of the subscripts on these components is to note that, starting from left to right, the first three subscripts in each of the three equations for the components of C are alwas cclic permutations of (,, ). Another wa to remember the order of combination of the unit vectors and the components of A and is to use this determinant: C = A = ˆ ŷ ẑ A A A MSN C. Direction of the Vector Product (a) rotational motion right hand (thumb is up, so C` is up) rotational ais ` A ` ` ` ` C = A rotational motion (screw backs out, so C` is up) rotational ais Figure 3. Two rules for finding the direction of C = A b: (a) the right hand rule; (b) the screw rule. (b) Epansion of this determinant leads to the same epression for C derived earlier. Show that the vector product of A = 5ˆ 2ŷ and = ˆ + ŷ + 3ẑ has these components: C = 6, C = 15, C =

7 MSN PS-1 MSN PS-2 PROLEM SUPPLEMENT Note: Problems 6 and 7 also occur in this module s Model Eam. current of 10 A and it is in a uniform magnetic field of = 0.50 T ˆ, calculate the torque about the -ais acting on the loop when θ = 60. Help: [S-5] 5.0 cm 1. Calculate the force on each of the five current-carring wire segments shown below, if the field is = 1.10 T ˆ, = 3.0 A, and d = 0.15 m. Consider each wire individuall. Help: [S-1] d d 8.0 cm d ` 4. A rectangular current loop, 4.0 cm b 8.0 cm and carring a current of 0.1 A, is suspended at a single point as shown. t is in a uniform horiontal magnetic field of magnitude 0.75 T. Find the magnitude of the torque acting on the loop when the normal to the plane of the loop makes an angle of 30 with respect to the magnetic field. 2. A section of a current-carring wire is fied so that it can slide up and down on two vertical metal guides as shown below. What magnetic field (magnitude and direction) is needed to prevent the sliding section from dropping and breaking the connection? The section is 0.30 m long, weighs 5 N, and has a current of 5.0 A passing through it. Help: [S-5] 0.30 m 3. A rectangular current loop, 5.0 cm b 8.0 cm, is fied on one side so that it rotates about the -ais as shown below. f it carries a 5. Calculate the magnitude of the maimum torque on a coil 3.0 cm b 5.0 cm, composed of 500 turns, when it carries a current of A in a uniform magnetic field of magnitude T. Help: [S-7] 6. A 500-turn square coil (2 cm on a side) in the - plane is in a magnetic field of magnitude = 0.16 T and direction ˆ. A current is passed through the coil, and it is observed that an eternal torque of N m ẑ is required to hold the coil in place. What are the magnitude and direction of? 13 14

8 MSN PS-3 MSN PS-4 7. A rectangular current loop is in a magnetic field of magnitude = 0.5 T and direction ˆ. The plane of the loop makes a 60 angle with the direction of the magnetic field. The dimensions of the loop and direction of current are as shown in the figure, and = 20 A. Calculate the forces on each of the four sides. Then calculate the torque, on the entire loop, about the -ais. 8 cm 8 cm 8 cm 60 rief Answers: 1. F1 = N ŷ; F 2 = 0; F 3 = N ŷ N ẑ = N (ŷ ẑ); F 4 = N ẑ; F 5 = N ŷ N ẑ = N (ŷ ẑ). 2. Magnetic field must have a horiontal component of 3.3 T into the page. Help: [S-2] 3. τ = N m ẑ 4. τ = N m 5. τ = N m 6. = 0.05 A. Observed from above, flows clockwise around the coil. 7. Top: F = 1.4 N ŷ Front: F = 0.8 N ẑ ottom: F = 1.4 N ŷ ack: F = 0.8 N ẑ t = N m ŷ 15 16

9 MSN AS-1 MSN AS-2 SPECAL ASSSTANCE SUPPLEMENT S-1 (from PS, problem 1) 1. Recall that we alwas use a right-handed coordinate sstem, i.e. ˆ ŷ = ẑ, so the positive -direction is into the page (awa from ou). Thus the cube pictured in Problem 1 is in the quadrant of 3-dimensional space where all coordinates are positive. Do not be put off b the orientation of the aes in the figure: aes can be shown in an orientation as long as the show a right-handed coordinate sstem. 2. To begin the problem, do as alwas: Decide on the relevant equation to use, then write down each vector quantit in that equation in terms of unit vectors and given quantities, using smbols to represent the given quantities. One wa to finish the problem is to then solve each component equation separatel. S-2 (from PS, problem 2) The sliding section of current-carring wire needs a magnetic force in the +-direction to cancel the weight of the wire. The direction of the current (and thus l) is in the +-direction. Since F = l and F must be mutuall perpendicular to l and, the onl possible choice for is in the positive or negative -direction. Take either choice and appl the right-hand rule to determine if it gives the correct direction for F. S-3 (from TX, 2a) The current in side a goes left so l a is in the ˆ direction. The magnetic field goes right so ˆ is in the +ˆ direction. The angle between the two is 180 so their vector product is ero (see this module s appendices): S-4 (from TX, 2a) The current in side b goes up so l b is in the +ŷ direction. The magnetic field goes right so ˆ is in the +ˆ direction. The angle between the two is 90 so their vector product is (see this module s appendices): F = L(ŷ ˆ) = L( ẑ). The direction ẑ is given b (see this module s appendices): ˆ ŷ = ẑ, so the direction +ẑ is out of the page and consequentl ẑ is into the page. S-5 (from PS, Problems 1, 2, 3) First do the eample in Sect. 2a, in ecruciating detail. When ou understand ever nuance of that eample, do the eample in Sect. 2b in similar detail. Then ou should be able to solve these problems on our own. S-6 (from TX, 2a) Recall that r b is the vector from the rotation ais to the point of application of the force. Now notice that the force on side b is in the opposite direction to the force on side d (remember?). Then the ais about which the loop rotates cuts down through the center of the loop, bisecting sides a and c [see Fig. (1)]. Then the vector from that central vertical ais out to side b is: r b = (W/2)ˆ [see Fig. (1)]. We alread found that F b = L( ẑ), so: τ b = r b F b = (W/2)(L)ˆ ( ẑ) = (W/2)(L)ˆ ẑ = +(W/2)(L)ŷ ( ˆ) (+ˆ) =

10 MSN AS-3 MSN ME-1 S-7 (from PS, Problem 5) The word 500 turns indicates that the continuous insulated wire was coiled around and around 500 times and then the whole set of 500 turns was glued together, keeping roughl the same shape as a single turn. The result is that 500 times as much current passes an one point on the loop as would pass that point if there was onl one turn of wire. f ou are having trouble finding the angle, look at the word maimum. MODEL EXAM 1. See Output Skills K1-K2 in this module s D Sheet. 2. A 500-turn square coil (2 cm on a side) in the - plane is in a magnetic field of magnitude = 0.16 T and direction ˆ. A current is passed through the coil, and it is observed that an eternal torque of N m ẑ is required to hold the coil in place. What are the magnitude and direction of? 3. 8 cm 8 cm 8 cm 60 A rectangular current loop is in a magnetic field of magnitude = 0.5 T and direction ˆ. The plane of the loop makes a 60 angle with the direction of the magnetic field. The dimensions of the loop and direction of current are as shown in the figure, and = 20 A. Calculate the forces on each of the four sides. Then calculate the torque, on the entire loop, about the -ais

11 MSN ME-2 rief Answers: 1. See this module s tet. 2. See Problem 6 in this module s Problem Supplement. 3. See Problem 7 in this module s Problem Supplement

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