MAGNETIC EFFECTS ON AND DUE TO CURRENT-CARRYING WIRES

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1 22 January _phys230_expt3.doc MAGNETIC EFFECTS ON AND DUE TO CURRENT-CARRYING WIRES OBJECTS To study the force exerted on a current-carrying wire in a magnetic field; To measure the magnetic field produced by a current-carrying wire; To determine the torque exerted on a current-carrying rectangular loop in a magnetic field. THEORY The magnetic force exerted on a straight wire of length L carrying current I in a magnetic field B is: F IL B ; F ILBsin B The magnetic field at a distance r from a long straight wire carrying current I has a magnitude of: B oi B 2 r Consider the current balance apparatus shown below: Figure 1 It consists of a movable coil and a pair of stationary coils. The three coils consist of ten turns of #20 enamelled, single-cotton-covered, magnet wire. The movable coil is pivoted on two knife edges so that it can rotate about a horizontal axis AA through these knife edges. Consider the wires at the left end of the apparatus as shown in Figure 1. The current I s in the top wire of the stationary coil is parallel to the current I m in the movable coil. Thus, the movable coil experiences an upward magnetic force of attraction toward the top of the stationary coil, causing a torque about the horizontal axis AA. Furthermore, the movable coil current I m is antiparallel to the current I s in the bottom part of the stationary coil, so there is a magnetic force of repulsion between these currents. This repulsion force is again upward on the movable coil thereby doubling the torque about the horizontal axis.

2 22 January _phys230_expt3.doc Examining the horizontal currents on the right side of the current balance shows that the right side of the movable coil experiences two downward magnetic forces, a repulsion from the antiparallel current in the top of the stationary coil and an attraction toward the parallel current in the bottom part of the stationary coil. The result is again a torque on the movable coil. The net result of the upward forces on the left end of the movable coil and the downward forces on the right end is a magnetic torque which causes the plane of the movable coil to rotate clockwise about the horizontal axis AA as seen from a front view of Figure 1. Attached to the movable coil are two rods carrying two sliding weights one large and one small. The sliding weights can be moved to provide gravitational torques which counteract the magnetic torque, and hence can be used to bring the movable coil to equilibrium, as explained in the following discussion. Consider the cross-sectional view of the current balance shown in Figure 2. On the left, the currents in the top stationary coil and the movable coil are parallel, so there is an upward magnetic force of attraction on the movable coil with magnitude given by F B = B s n m I m (1) where B s is the magnetic field, at the position of the movable coil, due to the current I s in the top part of the stationary coil, s/2 is the separation between the stationary and movable coils at equilibrium, I m is the current in the movable coil, n m is the number of turns (10) in the movable coil, and is the length of the left end of the movable coil that is parallel to the wires in the stationary coil. Also on the left side there is an upward force F B (identical to that of equation (1)) which arises due to the repulsion between the antiparallel currents in the movable coil and the bottom part of the stationary coil. The total clockwise torque on the movable coil due to the forces on the left side of the movable coil is thus: = 2F B d (2) where d is the distance between the supporting knife edge and the outer edge of the movable coil.

3 22 January _phys230_expt3.doc Because of the current directions, the right side of the movable coil experiences a net downward force of 2F B, and the torque due to this force is also clockwise with a magnitude of 2F B d. The total clockwise magnetic torque is thus mag = 4F B d (3) Substituting for F B yields: which can be re-arranged as: mag = 4(B s n m I m )d (4) mag = n m I m (2 d)(2b s ) (5) Note that 2 d is the area of the movable coil, A, and 2B s is the total magnetic field, B tot, acting on the ends of the movable coil: mag = n m I m AB tot (6) which is of the form B where = n m I m A is the magnetic moment of the movable coil. The clockwise magnetic torque can be counterbalanced by means of a gravitational torque provided by sliding aluminum weights. The current-direction switch on the apparatus is set so that when the current I is sent through the current balance, the end of the movable coil near the reference scale goes upward. This magnetic torque can be counterbalanced by a gravitational torque, SW, produced by the sliding large and small aluminum weights. SW = m L gr L + m S gr S (7) where r refers to the distances the masses have been moved from their initial positions and the subscripts L and S refer to large and small respectively. At equilibrium, SW = mag (8) EXPERIMENT FORCE ON A CURRENT-CARRYING WIRE IN A MAGNETIC FIELD (PASCO CURRENT BALANCE) PART A: Force versus Current 1. Set up the PASCO Current Balance equipment as shown in Figure 3, using Current Loop #38. Check that all 6 magnets are attached to the bottom of the pole pieces.

22 January 2013 4 2013_phys230_expt3.doc Figure 3 2. Turn the voltage and current controls of the DC power supply to their minimum settings and connect the circuit shown in Figure 4. Figure 4 3.

Press the Zero button on the electronic balance so that it reads 0.00 g. 4. Turn on the power supply. Turn the voltage control to mid-range. Adjust the current control until the ammeter reads 0.50 A.

4 22 January _phys230_expt3.doc Figure 3 2. Turn the voltage and current controls of the DC power supply to their minimum settings and connect the circuit shown in Figure 4. Figure 4 3. Position the main unit on the ring stand so that Current Loop #38 sits in the gap of the magnet assembly, centred between the pole pieces without quite touching the assembly. Press the Zero button on the electronic balance so that it reads 0.00 g. 4. Turn on the power supply. Turn the voltage control to mid-range. Adjust the current control until the ammeter reads 0.50 A. Observe the mass value reading from the electronic balance. If the reading is negative, remove the magnet assembly and turn it around so that the polarity of the magnetic field is reversed. Record the positive mass value reading. 5. Collect data in 0.50 A increments from 0.50 to 3.00 A. PART B: Force versus Magnetic Field 1. Turn off the power supply. Attach Current Loop #40 to the main unit. 2. Remove all but one of the magnets from the magnet assembly. The magnets can be removed one at a time by pressing the central magnet down against the support spring, and then sliding the magnet out of the assembly. 3. Centre the remaining magnet on the assembly.

22 January 2013 5 2013_phys230_expt3.doc 4. Place the magnet assembly on the balance and Zero the balance to 0.00 g. 5. Turn on the power supply, set the current to 3.

5 22 January _phys230_expt3.doc 4. Place the magnet assembly on the balance and Zero the balance to 0.00 g. 5. Turn on the power supply, set the current to 3.00 A, and record the mass value reading. Turn off the power supply after the reading is made. 6. For a constant current of 3.00 A, measure the mass value readings for 2, 3, and 4 magnets. PART C: Force versus Length of Wire 1. Put all six magnets in the magnet assembly. 2. For a constant current of 3.00 A, measure the mass value readings when using each of the six Current Loops. Current Loop Length (cm) SF SF SF SF SF SF PART D: Force versus Angle 1. Replace the Current Loop with the Current Balance Accessory as shown in Fig. 5. Place the wider gap alternate magnet assembly on the balance. Figure 5 2. Check that the accessory unit is close to but not touching the magnet assembly. Rotate the magnet assembly so that the magnetic field is approximately parallel to the wires on the bottom of the accessory coil, then set the red line to Zero the balance to 0.00 g. Set the current control to minimum, turn on the power supply, and set the current to 2.00 A.

6 22 January _phys230_expt3.doc 4. Record the mass value readings for angles from 0 to 90 in 10 increments. MAGNETIC FIELD PRODUCED BY A CURRENT-CARRYING WIRE (WOOD CURRENT BALANCE) 1. Carefully remove the movable coil from the apparatus. 2. Measure s, the average height of the stationary coils,, the average width of the movable coil, and 2d, the average length of the movable coil. Record m L and m S, the masses of the large and small aluminum weights. 3. Connect a jumper wire between the tabs on the posts on which the movable coil sits. 4. Connect the circuit shown in the following diagram: 5. Turn the voltage and current controls of the DC power supply to their minimum settings. Turn on the power supply. Turn the voltage control to mid-range. Adjust the current control until the ammeter reads 3.00 A. 6. Using the Hall effect gaussmeter, measure the magnetic field at various locations along the horizontal centre-line that is a distance of s/2 from the top or bottom of one of the stationary coils (the dashed line in the following diagram): Turn off the power supply when you have finished taking your measurements. TORQUE ON A CURRENT-CARRYING LOOP (WOOD CURRENT BALANCE) 1. Remove the jumper wire and re-mount the movable coil on the posts of the current balance. 2. Adjust the large and small aluminum weights so that they are at the 1 cm mark (closest to the central axis of the movable coil. 3. Balance the movable coil by adjusting the positions of the threaded weights on the brass rods until the pointer on the movable scale aligns with the reference mark on the apparatus. 4. Pass a current of 3.00 A through the current balance and adjust the reversing switch so that the pointer on the end of the movable coil moves upward relative to the reference mark. Counterbalance the resulting magnetic torque by moving the sliding aluminum weights until

7 22 January _phys230_expt3.doc the movable coil returns to equilibrium (as indicated by the pointer aligning with the reference mark). Record the final positions of the large and small aluminum weights. ANALYSIS FORCE ON A CURRENT-CARRYING WIRE IN A MAGNETIC FIELD (PASCO CURRENT BALANCE) Note that the force exerted on the Current Loop is directly proportional to the mass reading obtained on the electronic balance. Therefore, the shapes of the relationships of force versus current, magnetic field, length, and angle will be the same as the shapes of the relationships of mass reading versus current, magnetic field, length, and angle. 1. Plot mass reading versus current. 2. Plot mass reading versus # of magnets. 3. Plot mass reading versus length. 4. Plot mass reading versus sin where is the angle between the current-carrying wire and the magnetic field. 5. Discuss the shapes of your graphs in the context of F IL B B. MAGNETIC FIELD PRODUCED BY A CURRENT-CARRYING WIRE (WOOD CURRENT BALANCE) 1. Calculate B tot = 0 nsi s 2 2 s. Ignoring end effects (i.e. assuming that the stationary coil is 2 very long compared to the width of the movable coil) this is the magnetic field at the equilibrium position of the end of the movable coil due to the current I s in the stationary coil. Note that s/2 is the separation between the stationary and movable coils at equilibrium and n s is the number of turns in the stationary coils (10). Use g = 9.81 m/s 2, o = T m/a. Compare this calculated value for B tot with your measurements of the magnetic field along the centre-line of the stationary coil. TORQUE ON A CURRENT-CARRYING LOOP (WOOD CURRENT BALANCE) 1. Use the calculated value of B tot from above in equation (6) to calculate the theoretical magnetic torque acting on the movable coil. 2. Use equation (7) to calculate the gravitational torque acting on the movable coil when it is in equilibrium. 3. Compare the torques calculated in 1. and 2. and discuss.

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE

MAGNETIC FORCE ON A CURRENT-CARRYING WIRE Pre-Lab Questions Page 1. What is the SI unit for Magnetic Field? Name: Class: Roster Number: Instructor: 2. The magnetic field on a wire is 12.0 x 10 5 Gausses,