Problem Solving 4 Solutions: Magnetic Force, Torque, and Magnetic Moments

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 004 Problem Solving 4 Solutions: Magnetic Force, Torque, an Magnetic Moments OJECTIVES 1.

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.0 Spring 004 Problem Solving 4 Solutions: Magnetic Force, Torque, an Magnetic Moments OJECTIVES 1. To start with the magnetic force on a moving charge q an erive the force on a wire segment carrying current I.. To calculate the torque on a rectangular loop of current-carrying wire sitting in an ernal magnetic fiel. 3. To efine the magnetic ipole moment of a loop of current-carrying wire an write the torque on the loop in terms of that vector an the ernal magnetic fiel. REFERENCE: Sections , 8.0 Course Notes. Magnetic Force on a Straight Wire in an External Magnetic Fiel Consier a straight wire of length L an cross sectional area A which carries a current I. The current consists of moving positive charge. The wire is place in a constant ernal magnetic fiel. We want to calculate the force on this wire ue to the magnetic force acting on the charge carriers in the wire. We take an infinitesimal current element of length x in the ernal magnetic fiel as shown in Figure 1. Figure 1: Infinitesimal current element of length x in an ernal magnetic fiel. Solving4-1

2 In Figure 1, the charge carriers are assume to be positive for simplicity an move to the right with a rift velocity v. Suppose the ensity of positive charge carriers is ρ coulombs/cubic meter. Question 1 (Write your answer on the tear-sheet at the en!): What is the amount of charge q in our line segment in terms of ρ, A, an x? q = ρ ( volume) = ρ Ax Question (Write your answer on the tear-sheet at the en!): In a time t, all the charge in the element of length x = v t will cross the area A at the en of the element. Fin an expression for the total charge q that crosses the area A in this time t. x = v t, so q = ρav t Question 3 (Write your answer on the tear-sheet at the en!): Fin an expression for the current in the wire in terms of ρ, A, an v. Remember, the current in the wire is the amount of charge that flows pass a fixe point ivie by the time it takes for that amount of charge to flow past that point. q I =, an from above, q = ρavt, so I = ρav t Question 4 (Write your answer on the tear-sheet at the en!): Introuce an infinitesimal current element vector Is for a current carrying wire. This vector is tangent to the wire an points in the irection of the current, with s = x. Since the charge carriers are moving in an ernal magnetic fiel, there is a infinitesimal magnetic force F = qv. Using your answers from Question 1 an Question 3, you can show that F = qv = Is. In your answer on the tear-sheet, write own the algebraic steps you use to get this result. F = qv = ρaxv = ρav s, where we can make the last step because s an v are parallel to one another. Thus, using our expression for I above, F = I s, as esire. The total force on the wire is the sum of all these infinitesimal forces. This is an integral over the wire, Solving4-

3 F = F = I s wire wire Force on a Rectangular Current Loop in a Uniform Magnetic Fiel Given our result above, we now want to calculate the force on a loop of current-carrying wire sitting in a constant ernal magnetic fiel. Take a small rectangular square loop, with sies a an b, carrying a current I. The loop is place in a uniform magnetic fiel, which is in the plane of the loop an points to the right (Figure ). Let ˆn be the unit normal to the plane of the loop. We efine the irection of ˆn to be pointing in a irection efine by your right thumb when you curl the fingers of your right han in the irection of the current in the loop Figure : Current loop in an uniform magnetic fiel In the coorinate system shown above, = ˆi. The sies aligne along the z-axis are length a. The normal to the loop as we efine it above is nˆ = ˆj. Question 5 (Write your answer on the tear-sheet at the en!): Using the formula F = I s, calculate the force on sie 4 of the loop. Inicate both magnitue an wire irection. F = F = I s = I ( kˆz) ( ˆi) = I( kˆ ˆi) z =+ Iaj ˆ 4 sie4 sie4 sie4 sie4 Question 6 (Write your answer on the tear-sheet at the en!): Calculate the force on sie 3 of the loop. Inicate both magnitue an irection. Solving4-3

4 F = I ( ˆ i x) ( ˆ i ) = I( ˆ i ˆ i ) x =0 3 sie3 sie3 Question 7 (Write your answer on the tear-sheet at the en!): Calculate is the force on sie of the loop. F = I( kˆ ˆi) z = Iaj ˆ sie Question 8 (Write your answer on the tear-sheet at the en!): Calculate the force on sie 1 of the loop. F = I ˆix ( ˆi) = I( ˆi ˆi) x= 0 1 sie1 ( ) sie1 Question 9 (Write your answer on the tear-sheet at the en!): What is the total force on the loop (sum the forces on each sie)? The total force is zero. Torque on a Rectangular Current Loop in a Uniform Magnetic Fiel In Figure 3a (n page) we repeat Figure, an in Figure 3b, we show the loop as seen looking up the ˆk axis (the ˆk vector is into the page in Figure 3b). In orer to calculate the torque on sie 4 about the center of the loop, it is sufficient to consier the entire force on sie 4 as acting at the mipoint of the sie. The torque on sie 4 is given by τ = r F, where r 4 is the vector from the center of the loop to the mipoint of the sie an F 4 is the force we calculate in Question 5 above. Question 10 (Write your answer on the tear-sheet at the en!): What is the torque τ 4 about the center of the loop ue to the force acting on sie 4? Inicate its magnitue an irection. b ˆ 1 τ4 = r4 F4 = ( Ia ˆ) = Iba ˆ i j k Solving4-4

5 Figure 3: (a) A repeat of Figure ; (b) The loop as seen looking up the ˆk axis Question 11 (Write your answer on the tear-sheet at the en!): What is the torque τ about the center of the loop ue to the force acting on sie? b ˆ 1 τ = r F = ( Ia ˆ) = Iba ˆ i j k Question 1 (Write your answer on the tear-sheet at the en!): Why is the torque on sies 1 an 3 zero? ecause the forces on those sies are zero. Question 13 (Write your answer on the tear-sheet at the en!): What is the total torque τ = τ + τ on the loop? 4 τ = Iabkˆ Question 14 (Write your answer on the tear-sheet at the en!): What effect will this torque have on the loop? A torque in the kˆ irection will cause the loop to rotate counterclockwise when looking own the kˆ axis. Put the thumb of your right han along the torque vector, an that torque will cause the object on which it acts to rotate in the irection of your fingers. Solving4-5

6 Magnetic Dipole Moment The magnetic ipole moment vector µ of a planar current loop is efine as follows. For a planar loop, µ is efine in terms of the current I, the area A of the loop, an the unit normal ˆn to the plane of the loop, µ IA= IAˆ n The normal ˆn points in a irection efine by your right thumb when you curl the fingers of your right han in the irection of the current in the loop. The torque on a planar current loop or any magnetic ipole in a uniform magnetic fiel can then be shown to be τ magnetic = µ. This result mirrors our earlier result for electric ipoles. When we place an electric ipole p in a uniform electric fiel E, we foun that the torque on the electric ipole was given by τ = p E electric Question 15 (Write your answer on the tear-sheet at the en!): Calculate the torque on the planar current loop shown in Figure, using the efinitions an expressions given above. Does this agree with the result you foun in Question 13 above? The magnetic moment vector is µ IA= Iabˆj. The torque is therefore τ = µ = ( Iabˆj) ( ˆi) =+ Iabkˆ, magnetic. which agrees with our result above. Problem: A square loop of wire, of length on each sie, pivots about an axis AA' that correspons to a horizontal sie of the square, as shown in Figure 4. A magnetic fiel of magnitue is irecte vertically ownwar, an uniformly fills the region in the vicinity of the loop. The current I in the loop moves counterclockwise as viewe from above. Solving4-6

7 Figure 4: A current loop in an ernal fiel. Question 16 (Write your answer on the tear-sheet at the en!): Calculate the magnitue of the torque on this loop of wire in terms of the quantities given, using our expressions above. The magnetic ipole moment of this loop is its area times the current in it times the number µ = µ of turns, so Il, with the vector perpenicular to the plane of the loop an right-hane with respect to the current in the loop. With the current flow counterclockwise as viewe from the top, the vector µ is up an to the right. That vector crosse into the ownwar is into the plane of the paper, that is, µ is into the page. The magnitue of the torque in terms of the angle shown is µ = µ cosθ = Il cos θ. This is because the angle between the two vectors is π π θ, an sin θ = cos θ. Question 17 (Write your answer on the tear-sheet at the en!): Does the sense of this torque make the coil pivot so that the angle θ increases or ecreases? Put the thumb of your right han along the irection of the torque, an that torque will make the object on which it is exerte rotate in the irection your fingers curl. For this case this will be clockwise because the torque is into the page. Thus, the angle θ ecreases. Solving4-7

Problem Solving 6: Magnetic Force & Torque

OBJECTIVES MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Problem Solving 6: Magnetic Force & Torque 1. To look at the behavior of a charged particle in a uniform magnetic field by studying