# Experiment #9, Magnetic Forces Using the Current Balance

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1 Physics Fall Experiment #9 1 Experiment #9, Magnetic Forces Using the Current Balance 1 Purpose 1. To demonstrate and measure the magnetic forces between current carrying wires. 2. To verify the theoretical predictions of the force between parallel current carrying wires and to utilize this result to obtain a value for the magnetic permeability constant, μ 0. 2 Introduction Fig. (1) shows a current balance which consists of two parallel rods. One of the rods forms one side of a rectangular loop that is balanced on knife edges and the other stationary rod is mounted directly under one of the sides of the rectangular loop. The loop pivots freely on knife edges and is balanced by a counterweight mounted on the loop. The equilibrium position of the upper rod with respect to the lower rod is set by changing the position of the counterweight. A mirror, mounted on the loop, is used to reflect a beam of helium-neon laser light for accurately determining the position of the upper rod. A tiny tray is mounted on the upper rod for placing very small weights on it. Figure 1. The Current Balance. A DC power supply is connected to both rods so that they carry equal but oppositely directed currents which induce a repulsive force between them. This repulsive force causes the pivoting of the upper rod. Small weights placed on the upper rod (tray) disturb the equilibrium position which is restored by passing a current through the rods. The value of these small weights can be used to determine the permeability constant of free space μ 0 /4π. The following section explains three types of forces encountered in this experiment. 2.1 Magnetic and Gravitational Forces Magnetic force of repulsion (or attraction) on the top rod. Consider the current configuration shown in Fig. (2). Two parallel rods of length L, separated by a center-to-center distance R, carry equal currents I in opposite directions. If the rods are oriented parallel to the x-direction, the current passing through the lower rod produces a magnetic field B z in the z-direction at the position of the upper rod. This magnetic field interacts with the current in the upper rod and causes a repulsive force F y on the upper rod in the y-direction. This repulsive force pivots the rectangular loop about the knife edges. If the length L is large compared to their separation R, the magnitude of the repulsive force is given by: L Fy = ( )( )I 2π R 2 I BLI; B ( )( ) (1) 2π R Review the derivation of Eq. (1) in your textbook. The SI units for B is the Tesla T (N/A-m).

2 Physics Fall Experiment #9 2 Note: 1. The fact that the current in each rod is in opposite directions is the reason for the force on the upper rod to be repulsive, as indicated in Fig. (2a). If the current in each rod is in the same direction, the magnetic force will be attractive as indicated in Fig. (2b). 2. The current in the upper rod exerts an equal and opposite magnetic force (down) on the lower rod, in agreement with Newton's 3rd law of interaction, i.e., of action and reaction. Because the lower rod is fixed, it does not move under the application of this force. 3. The magnitude of the magnetic force is directly proportional to the length L of the upper rod and inversely proportional to their separation distance R. 4. As the force is proportional to the square of the current, reversing the direction of the current in both rods will not reverse the direction of the force, i.e., the force will not become attractive Magnetic force on the upper rod due to the presence of the earth's magnetic field. The earth's magnetic field B e will exert a force F e on a conducting rod of length L and carrying a current I. The x, y and z components of force F e are related to I, L and the x, y and z components of B e in the following way: F e,x = 0 F e,y = ILB e,z (2) F e,z = ILB e,y In derivation of the above equations, it is assumed that the x-axis of the rectangular coordinate system is oriented in the direction of the current flow in the top rod and the y-axis is along the vertical. See Fig. (3). B e will include other local magnetic fields as well as the earth s. There are dampening magnets on the device. Figure 3: Magnetic force due to earth s magnetic field.

3 Physics Fall Experiment #9 3 Please note: 1. The y-component of the force F e,y can only pivot the loop while the z-component of force F e,z can only translate the loop. Thus, only the y-component of the force is responsible for the pivoting of the loop. 2. When the direction of the current I is reversed, the direction of the force on the upper rod by the earth's magnetic field reverses. This is in contrast to no change in the direction of the (repulsive) magnetic force on the upper rod due to the current in the lower rod. See section Gravitational force When small weights are placed in the tray on the upper rod, the gravitational force W acts on the upper rod in the vertically down direction. The force W in term of the weight's mass m is given by: 2.2 Net force on the upper rod of the current balance. F y = W = mg (3) The three forces discussed in Sections 2.1.1, and act simultaneously on the upper rod in y-direction and cause it to pivot. The net force on the rod is obtained by adding these three forces given by Eqs. (1), (2) and (3). Thus, the net force that causes the pivoting of the upper rod is given by: F net,y = μ 2π L R o 2 ( )( )I ILB e,z W (4) 2.3 Current balance operation With no weights added and no current in the rod, the equilibrium position of the current balance is noted by marking the spot made by the beam of light from a laser on a screen. When a known weight is added to the tray the light spot on the screen will shift down to a new position. The current I 1 through the rods is then increased until the light spot is restored to its equilibrium position. When this happens, the net force on the upper rod is zero. Applying the condition, F net,y = 0, to Eq. (4), one obtains the first of the two balancing equations: W = L 2 ( )( )I1 I 1 LB e,z (5) 2π R If the direction (not the magnitude) of the current is reversed, there will be a shift in the position of the light spot. (This is so because the force due to the earth's magnetic field reverses its direction but other forces maintain their direction.) Thus, a current of different magnitude I 2 is needed to restore the light spot to its equilibrium position. When this is done, the net force on the upper rod is again zero. Thus, one obtains the second of the two balancing equations: L 2 W = ( )( )I2 I 2 LB e,z (6) 2π R Eqs. (5) and (6) can be rearranged to obtain the following two equations: B e,z = ( ) (I 2 - I 1 ) (7) 2πR

4 Physics Fall Experiment #9 4 L W = ( ) (I 1 I 2 ) (8) 2πR Eq. (7) is obtained by subtracting Eq. (6) from Eq. (5) and canceling the common factor (I 1 + I 2 ), and then rearranging the equation. Eq. (8) is obtained by first multiplying Eq. (5) by I 2 and Eq. (6) by I 1 and then adding them together, canceling the common factor (I 1 + I 2 ), and then rearranging the equation. Note that the currents I 1 and I 2 in the above equations are the magnitude of the forward and the reverse currents. Eqs. (7) and (8) are the working equations for this experiment. The experimental method consists of determining two balancing currents I 1 and I 2 for each weight W used. The data is then analyzed by plotting a graph of weights W (N) versus the product of balancing currents I 1 I 2 (A 2 ). According to Eq. (8), this graph should be a straight line whose intercept should be zero and whose slope S is given by: L Slope S = ( ) 2πR with units of A N 2 (9) Experimental Apparatus and Procedure 3.1 Apparatus 1. Welch Current Balance 2. Hampden variable DC power supply 3. Digital ammeter and voltmeter 4. DPDT switch (double pole/double throw) 5. Helium Neon Laser 6. Meter stick 7. A set of milligram standard masses 8. Rheostat and Various wires The Welch current balance consists of a pivoted, rectangle-shaped assembly of current-carrying rectangular loop which pivots about the balance axis. One side of the loop is parallel to a fixed rod below it. See Fig. (1). With rods wired in series, the currents flowing in them will be in opposite directions and the upper rod will experience a repelling force which causes the balance to pivot upwards. See Fig. (2). The orientation of the balance is monitored by the position of a laser beam reflected from a mirror attached to the balance frame [(Fig. (1)] to a screen placed several meters away from the balance [Fig. (4)]. When the balance rotates downward, the laser beam spot on the screen moves downward and vice versa. See Fig. (4) Figure 4: Beam of light reflected from a mirror.

6 Physics Fall Experiment #9 6 a. The digital current meter indicates reverse current I 2 with a negative sign. b. The balance pivots upward (not downward) indicating that the force on the upper rod is repulsive. c. The light spot on the screen moves upward. 7. Turn down the current control knob to the zero position and observe the light spot return to its equilibrium position. Switch the DPDT to its center off-position Measurements of currents I 1 and I 2 as a function of weight W placed on the upper rod. 1. Use the 6-column data table. 2. With the DPDT switch in its center off-position mark the equilibrium position of the light spot, i.e., carefully draw a pencil line around the spot and darken it. Also mark the vertical position of the top of the light spot with one horizontal lines. Check that the spot returns to your marked position by turning the power to the circuit slowly up and down a few times. From now on, the equilibrium position of the current balance refers to this marked position of the light spot. 3. a. Gently place the 10 milligram mass on the tray using the tweezers. The current balance will pivot down and the light spot will move vertically down from the equilibrium position. b. Rebalance the current balance by increasing the current from zero (in the arbitrarily designated "forward" position of the DPDT switch) until the laser spot has risen back to the equilibrium position. Record the forward balancing current I 1. c. Turn the current back to zero. Reverse the DPDT switch, thus reversing the direction of current flow in both rods. Rebalance the current balance by increasing the current until the light spot returns to the equilibrium position. Record the "reversed" balancing current I 2. Note that the digital ammeter indicates a negative sign before the displayed digits. 4. Gently add another 10 milligram mass to the tray, making the total mass on the tray to be 20 milligrams. Repeat Step Repeat Step 3 when the total mass on the tray is 30, 40, 50, 60, 70, 80, 90, and 100 milligrams. Caution: Once a weight is placed in the tray, do not remove any weights between steps in order to avoid jarring the current balance. 6. Turn off the Hampden power supply. Do not remove the weights. 7. Add the 1g weight. Observe that the upper rod touches the lower rod. Carefully draw a horizontal pencil line across the top edge of the light spot on the screen. Measure and record the distance y between the top and bottom horizontal line of the marks corresponding to the equilibrium and the 'clamped' positions of the light spot s top edges. 8. Additional Measurements: Make the following measurements necessary to obtain the value of the separation distance R between the upper and the lower rod as well as the value of the length L of the upper rod. Use a metric ruler to make these measurements. a. Measure the distance X from the pivot axis to the screen [Fig. (6)].

9 Physics Fall Experiment #9 9 Appendix A Derivation of Equation (10): Figure A1 indicates two positions of the upper rod; one position when the upper rod is at its equilibrium position and the other when the upper rod is touching the lower rod. Figure 1A: Indicating the relationship between the defection of the upper rod and the reflected laser beam. When the upper rod moves from its equilibrium position to the position when it touches the lower rod, the displacement of the upper rod is given by ( R d ) where R is the center to center separation of the upper rod with respect to the lower rod and d is the diameter of each rod. When the upper rod touches the lower rod, it pivots by an angle α and the laser beam reflected from the mirror deflects by an angle δ. When the angle α is small, it is related to (R- d) and x by: R - d = ( ) (13) x Similarly, the angle δ is related to the deflection y of the laser beam on the screen placed a distance X from the pivot point in the following way: y = ( ) (14) X It is the simple matter to show that the angle δ is twice the angle α for the laser spectral reflection, i.e.,: Inserting Eqs. (14) and (15) in (16) and then by rearranging, one obtains: yx R = d + ( ) 2X (16)

10 Physics Fall Experiment #9 10 Appendix B Data Sheet Caution: Please follow the procedure in the lab handout. Note the use of absolute values. No values in the table below should be negative when entered. Mass * I 1 I 2 I I 1 I 2 Weight * kg Amp Amp Amp Amp 2 N + Forward - Reverse x x x x x x x x x x * 1 milligram = 10-6 kg; W = mg, g = m/s 2 Additional Data: 1. Displacement of the laser light spot between equilibrium and clamped position of the current balance: y = cm 2. Distance from the pivot axis to the screen: X = cm 3. Distance from the pivot axis to the center of the upper rod: x = cm 4. Length of the current carrying rod: L = cm 5. Diameter of the upper rod: d = cm Error y = y = cm Error X = X = cm

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