Magnetic Force | Physics Homework Help Online

A popular demonstration in physics is to show the force between current carrying wires. Usually it’s expressed like this. Parallel wires with currents flowing in the same direction, display an attractive force. This isn’t quite right; the wires are not attracted to each other, like charges in Coulomb’s Law, there’s an intermediary here.

A current is the rate at which charges flow past a given point and moving charges establish magnetic fields in the space around them. For the wires in the demonstration, the magnetic field is circular and concentric on the wire in a direction determined by a right hand rule. Its magnitude can be approximated using Ampere’s Law.

A moving charge, the current, in the wire running parallel to the first wire moves through this magnetic field and experiences a force. The magnitude of the force on a given charge is given by. The Lorentz Force Law, its direction by another right hand rule.

Where q is the charge and v is its velocity. In the case of our demonstration the first term can be ignored and since the current, I, is charge/second and v is distance/second, the qv term can be rewritten as IL, giving us…

 

Where I’ve expanded the cross product, but since our wires, and their currents, are perpendicular to the magnetic field, θ is 90° and sinθ is 1. This gives us the model we’re going to test.

 

 

EQUIPMENT

Our setup was somewhat different from the system described in the introduction.

  • We generated a magnetic field using horseshoe magnets rather than a second wire. This can be justified since the charges in the current don’t care how or why the magnetic field exists; only that it does.
  • Also we measured the force on the magnets, rather than on the current carrying wire, which we can justified using Newton’s third law.

PROCEDURE 1

  • Fixed the length of the wire (SF 42) and the magnetic field, varying the current.
  • Measured the mass as the current changes.

DATA

MASS READINGS (kg)FORCE (N)CURRENT (A)
0.000240.002350.312
0.000290.002840.382
0.000330.003230.453
0.000350.003430.500
0.000400.003920.570
0.000440.004310.640
0.000480.004700.712
0.000500.004900.760
0.000540.005290.832
0.000580.005690.903

 

Force is calculate using Newton’s second law F=ma where a is our gravitational force which was 9.802 m/s2.

 

ANALYSIS

Magnetic field strength is calculated from the slope of the graph.

Slope = change in y/ change in x

Taking the two points (0.903, 0.00569) and (0.321, 0.00235)

Slope = (0.00569 – 0.00235) / (0.903-0.312)

= 0.00334/0.591

= 0.00565 T

Given Magnetic field strength B = 0.00565T and current = 3A force is calculated as follows;

HINT: SF 42 is 8.0cm which is equal to 0.08m

F = 3A * 0.08* 0.00565T

= 0.001356N

The intercept should not be set to 0.

 

PROCEDURE 2

  • Fixed the length of the wire (SF 42) and the current, varying the magnetic field.
  • Measured the mass as I varied the number of magnets.

DATA

MASS READINGS (kg)FORCE (N)MAGNETIC FIELD STRENGTH, B (T)
0.000150.001470.0580
0.000220.002160.0857
0.000170.001670.0656
0.000190.001860.0727
0.000160.001570.0609

 

Calculations;

  1. mass 0.00015 kg

F = ma

= 0.00015kg * 9.802m/s2

= 0.00147N

Magnetic field strength;

B = F/IL

L = 8.0 cm or 0.08m

B = 0.00147N/(0.317A*0.08m)

= 0.0580 T

 

  1. mass 0.00022 kg

F = ma

= 0.00022kg * 9.802m/s2

= 0.00216N

Magnetic field strength;

B = F/IL

L = 8.0 cm or 0.08m

B = 0.00216N/(0.315A*0.08m)

= 0.0857 T

 

  1. mass 0.00017 kg

F = ma

= 0.00017kg * 9.802m/s2

= 0.00167N

Magnetic field strength;

B = F/IL

L = 8.0 cm or 0.08m

B = 0.00167/N(0.318A*0.08m)

= 0.0656 T

 

 

  1. mass 0.00019 kg

F = ma

= 0.00019kg * 9.802m/s2

= 0.00186N

Magnetic field strength;

B = F/IL

L = 8.0 cm or 0.08m

B = 0.00186N/(0.320A*0.08m)

= 0.0727 T

 

  1. mass 0.00016 kg

F = ma

= 0.00016kg * 9.802m/s2

= 0.00157N

Magnetic field strength;

B = F/IL

L = 8.0 cm or 0.08m

B = 0.00157N/(0.322A*0.08m)

= 0.0609 T

 

From the graph plotted above, the force increases gradually until it reaches a maximum magnetic field strength, B of 0.857T and then it reduces again till it reaches a B of 0.656T. This force then increases until it reaches a B of 0.727T and then lastly reduces to a magnetic field strength of 0.0609T. I calculated the force using Newton’s second law F=ma since we had the mass and the acceleration. We then used the equation of magnetic field strength to calculate it and then plotted force against magnetic field strength. This was important to study the relationship between force and the magnetic field strength.

 

 

 

PROCEDURE 3

  • Fixed the current and the magnetic field, 6 magnets.
  • Measured the mass as I changed circuit boards.

DATA

CIRCUIT BOARDMASS READINGS (kg)FORCE (N)WIRE LENGTH (m)
SF 420.000390.003820.08
SF 410.000240.002350.06
SF 380.000260.002550.04
SF 390.000190.001860.03
SF 370.000120.001170.02
SF 400.000040.000390.01

 

Force is calculated using Newton’s second law equation F=ma;

  1. F = ma

m = 0.00039kg

a = 9.802 m/s2

         F = 0.00039kg*9.802m/s2

= 0.00382 N

  1. F = ma

m = 0.00024kg

a = 9.802 m/s2

                 F = 0.00024kg*9.802m/s2 = 0.00235 N

 

  1. F = ma

m = 0.00026kg

a = 9.802 m/s2

 F = 0.00026kg*9.802m/s2 = 0.00255 N

  1. F = ma

m = 0.00019kg

a = 9.802 m/s2

 F = 0.00019kg*9.802m/s2 = 0.00186 N

  1. F = ma

m = 0.00012kg

a = 9.802 m/s2

 F = 0.00012kg*9.802m/s2 = 0.00117 N

  1. F = ma

m = 0.00004kg

a = 9.802 m/s2

 F = 0.00004kg*9.802m/s2 = 0.00039 N

Magnetic field strength is calculated from the slope of the plot above. The slope is calculated by change in y divided by change in x. Taking point (0.08, 0.00382) and (0.02, 0.00117) the slope is calculated as below:

Magnetic field strength B = (0.00382-0.00117)/ (0.08-0.02)

= 0.0442 T

CONCLUSION

My results display the relationship discussed in the introduction whereby an attractive force was displayed when current in parallel wires was flowing in the same direction, magnetic fields were established around currents flowing past a given point and the force depends on the length of the wire.

In the first procedure the calculated magnetic field strength was 0.00565T and in the last procedure the calculated magnetic field strength was 0.0609. The two values do no agree. % difference is calculated below;

% = {0.0609-0.00565/ [(0.0609+0.00565)/2]}*100

= (0.05525/0.033275)*100

= 1.66

This demonstration work as follows; power comes from the power supply to the current meter through wire to the meter and back to the power supply. The current is controlled by increasing current in a single wire or keep constant for a series of wires. Magnetic field is generated using magnets aligned with red on one side and white on the other side vaguely horse shoe in shape. Newton’s third law justifies measuring the force on the magnets instead on the sire because two magnets repelling each other or attracting each other produces a force between them which is the same on each magnet and points on opposite directions on each of them. The direction of the current effect the force on the magnets as stated from the right hand rule which says that magnetic field lines which are produced by a current currying wire will be oriented in same direction as curled fingers of a person’s right hand with thumb pointing in the direction the current is flowing.

Sources of experimental error could be due to difficulty in making the current constant. Errors can also be caused by the person taking values immediately the current is switched on which not the case since one should wait till the current stops varying and settles to a constant value. In the last experiment, it was a little bit difficult to obtain the values since the SF 40 has the least wire length. Magnets do not liked to be aligned in horse shoe shape because the poles are much closer together creating a more direct path for flux lines to travel between them making the field to be concentrated around the poles.

 

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Magnetic Force | Physics Homework Help Online . (2021, December 27). Essay Writing . Retrieved August 17, 2022, from https://www.essay-writing.com/samples/magnetic-force/
“ Magnetic Force | Physics Homework Help Online .” Essay Writing , 27 Dec. 2021, www.essay-writing.com/samples/magnetic-force/
Magnetic Force | Physics Homework Help Online . [online]. Available at: <https://www.essay-writing.com/samples/magnetic-force/> [Accessed 17 Aug. 2022].
Magnetic Force | Physics Homework Help Online [Internet]. Essay Writing . 2021 Dec 27 [cited 2022 Aug 17]. Available from: https://www.essay-writing.com/samples/magnetic-force/
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