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Posted: Sun Apr 17, 2022 3:55 pm
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A 10 mg mass is placed on the upper rod tray of a current balance and is brought to equilibrium when the forward balancing current is 14 = 3.47 A and the reversed current is 12 = 2.99 A. When 100.00 mg is placed on the tray, the forward balancing current is I1 = 9.76 A and the reverse current is 12 = 9.50 A. 1. Using this information, find the slope of the two (W,1,12) pairs (use g = 9.8035 m/s2) and report it to three decimal places. 2. Then, use R = 0.476 cm and L = 26.50 cm, find the magnetic permeability of free space (10/471) and report it to three significant figures. 3. Compare your value to the accepted value using a percent difference calculation and report it to two significant figures. I
Slope = Mo/4n = %diff=
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, in the z-direction at the position of the upper rod. This magnetic field interacts with the current in the upper rod and auses a repulsive force on the upper rod in the y- direction. This repulsive force pivots the rectangular loop about the knife edges. If the length Lis large compared to their separation R, the magnitude of the repulsive force is given by: Fy = 2 = BLI: (1) Review the derivation of Eq. (1) in your textbook. The SI unit for B is the Tesla, 7 (Nam) The magnetic permeability of free space, Ho is what we seek to determine in our experiment. The accepted value of Ho is defined as: B 2 R N Ho = 4 x 10- We will seek a less cumbersome version HO = 1 x 10-7 476 N 42
Magnetic force on the upper rod due to the presence of the earth's magnetic field. The earth's magnetic field, Be, will exert a force, F., on a conducting rod of length L carrying a current I. The components of the force, Fex, Fey, Fez, are related to current, length, and the components of B, in the following way: Fex = 0 Fey = EILB ( (2) Foz = EILBey 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). Be will include other local magnetic fields as well as the earth's (the magnetic field due to the dampening magnets on the device, for example).
! 30 1 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 vertical direction. The force, in terms of the added mass, mis given by: Fy = W = mg (3) Net force on the upper rod of the current balance The three forces discussed above act simultaneously on the upper rod in the y direction and cause it to pivot. Thus, the net force that causes the pivoting of the upper rod is given by the sum of these force of electric attraction/repulsion, force of the local magnetic field, and force of gravity: 3 I UMass Boston Physics 182 Spring 2022 Frety B: -W 2x
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 lasct 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, Is through the rods is the increased until the light spot is restored to its equilibrium position. At this time, the net force on the upper rod is zero. Applying the condition, Frety=0. to Eq. (4), and solving for W.we obtain the first of the two balancing equations whl. Bu (5) 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 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 is needed to restore the light spot to its equilibrium position. When this new current is applied the net force on the upper rod is again rero. Thes, one obtain the second of the two balancing equations 1 W- (6) Eq. (5) and (6) can be rearranged to obtain the following two equations B-30,- (7) M 1 18 Eq. (T) is obtained by wabtracting Eq. (6) from 1 (5) and can them
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Slope = Mo/4n = %diff=
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, in the z-direction at the position of the upper rod. This magnetic field interacts with the current in the upper rod and auses a repulsive force on the upper rod in the y- direction. This repulsive force pivots the rectangular loop about the knife edges. If the length Lis large compared to their separation R, the magnitude of the repulsive force is given by: Fy = 2 = BLI: (1) Review the derivation of Eq. (1) in your textbook. The SI unit for B is the Tesla, 7 (Nam) The magnetic permeability of free space, Ho is what we seek to determine in our experiment. The accepted value of Ho is defined as: B 2 R N Ho = 4 x 10- We will seek a less cumbersome version HO = 1 x 10-7 476 N 42
Magnetic force on the upper rod due to the presence of the earth's magnetic field. The earth's magnetic field, Be, will exert a force, F., on a conducting rod of length L carrying a current I. The components of the force, Fex, Fey, Fez, are related to current, length, and the components of B, in the following way: Fex = 0 Fey = EILB ( (2) Foz = EILBey 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). Be will include other local magnetic fields as well as the earth's (the magnetic field due to the dampening magnets on the device, for example).
! 30 1 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 vertical direction. The force, in terms of the added mass, mis given by: Fy = W = mg (3) Net force on the upper rod of the current balance The three forces discussed above act simultaneously on the upper rod in the y direction and cause it to pivot. Thus, the net force that causes the pivoting of the upper rod is given by the sum of these force of electric attraction/repulsion, force of the local magnetic field, and force of gravity: 3 I UMass Boston Physics 182 Spring 2022 Frety B: -W 2x
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 lasct 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, Is through the rods is the increased until the light spot is restored to its equilibrium position. At this time, the net force on the upper rod is zero. Applying the condition, Frety=0. to Eq. (4), and solving for W.we obtain the first of the two balancing equations whl. Bu (5) 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 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 is needed to restore the light spot to its equilibrium position. When this new current is applied the net force on the upper rod is again rero. Thes, one obtain the second of the two balancing equations 1 W- (6) Eq. (5) and (6) can be rearranged to obtain the following two equations B-30,- (7) M 1 18 Eq. (T) is obtained by wabtracting Eq. (6) from 1 (5) and can them