Physics Lab Report 1
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1
Physics Lab Report 1
Magnetic Fields
Course: PCS130
Section: 09
Teaching Assistant: Ary Sakafish
Instructor: Tetyana Antimirova
Lab Partners: Ahmed Abdul Lathif, Rafia Kibria
January 17, 2023
2
Table of Contents
Introduction ……………………………………………… 3
Theory …………………………………………………… 4
Procedure ………………………………………………
....
5-7
Results and Calculations …………………………………. 8-12
Discussion and Conclusion ………………………………. 13-14
References ………………………………………………… 15
3
Introduction
A Magnetic field is a vector field that describes the influence magnetism has on moving
electric charges. Electric charges that are in motion create magnetic fields due to the electrical
current flow in a conductor. The strength and direction of the magnetic field depend on how
much current is flowing and the direction of its flow. The magnetic fields created can be
calculated by the
Biouts-Savart Law
known as
=
. In this experiment, we studied
??
µ0
4π
?(?𝐿
×𝑟
)
𝑟
2
the magnetic fields generated by an electric current passing through the conductor which in this
case was a coil of a conductive metal. In discovery, we found the shape and size of the coils to
impact the magnetic field that was generated, which showed us the relationship of the magnetic
field and the electrical current.
We also conducted this experiment to understand the relations the distance and the
strength of the magnetic field at those specific points had by using two coils of wires and a 10.5
cm radius. To complete this experiment with success, three parts such as finding the
magnetic
field in the center of the coil, the magnetic field along the central axis of a single coil, and the
magnetic field along the central axis of a helmholtz coil
were to be conducted.
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4
Theory
The Biot-Savart Law is used to calculate magnetic fields created by the electric current:
??
→
=
µ
0
4π
?(?𝐿
→
𝑥 𝑟
^
)
𝑟
2
According to the equation, a very small magnetic current
is generated by a current-carrying
??
→
( ) conductor of a small cross-section
at a point
distance away from it. The constant
is
?
?𝐿
→
𝑟
µ
0
the permeability of vacuum.
We can use the Biot-Savart Law to find the magnetic field due to a current-carrying loop of wire.
Consider a circular loop that has a radius R, carries a current I, and lies in the
xz
-plane:
5
Procedure
Part (I) Magnetic Field at the center of a single coil
1.
Connect the Magnetic Field Sensor to CH-1 of the LabPro computer interface and open
logger pro. Don’t forget to calibrate the field sensor.
2.
With the power supply
off
, connect a single coil using the white plugs to the power
supply and set it to the
6.4mT
range.
3.
Position the probe so that it is in the same plane as the coil. The probe will stay in this
position for
this part of the lab.
4.
Turn on the power supply and set the electric current to
0.4A.
Do not leave the power
supply on for too long of a period to avoid your electric current value being affected.
5.
In the LoggerPro interface, the top graph displays the
magnetic field vs. time
. Select
Analyze
→
For statistics.
6.
Record the magnetic field results, as well as the electrical current in the excel sheet and
assign an uncertainty to your magnetic field value based on the fluctuations observed.
7.
Measure the magnetic field every
0.2 A
up to a
maximum
of
2 A
. Record the results
and electric current, I, each time.
8.
Determine the relationship between the magnetic field of a single coil, Bcoil, and electric
current, I by plotting your results and applying a linear fit.
6
Part (II) Magnetic Field along the central axis of a single coil
1.
With the power supply
Off
, zero the magnetic field sensor while it is still at the center of
the coil and with a single coil still connected, set the power supply so that the current is 2
A when it is on.
2.
With the power supply on, move the probe and ruler away from the coil until it reaches
approximately
20%
of the maximum magnetic field strength. Switch to the second sheet
(B vs. z).
3.
Record measurements of the magnetic field strength and position along the central axis
in increments of 2 cm until you reach the same magnetic field strength on the opposite
side of the coil.
4.
You should record an uncertainty value for the magnetic field sensor’s position and test a
fit by creating a set of data from the
theoretical equation:
?
?𝑜𝑖?(?, 𝑧) = ?
?
2
((𝑍+?)
2
+?)
2
)
3/2
𝑧
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7
Part (III) Magnetic Field along the central axis of a Helmholtz coil
1.
With the power supply
Off,
zero the magnetic field sensor and position two coils within
the rectangular outlines on the base. Coils must be parallel with one another.
2.
Connect two coils in series using the white plugs such that the magnetic fields of each
coils point in the same direction. Verify that the coils are correctly configured to one
another by
right hand rule
for coils.
3.
Set the power supply to output
1 A
of current.
4.
Create a new sheet in your Excel file called
Helmholtz Coil.
5.
Measure and record the magnetic field across the Helmholtz coil in steps of 2 cm all the
way through the two coil system.
6.
Write the equation for the magnetic field of a two coil system with the coils
R distance
apart.
7.
Fit your Helmholtz data using the equation you found above and 2-3 fitting coefficients.
8.
Record relevant values such as the radius of the loops, number of turns, and
distance between the two coils
9.
Lastly, tidy up your workstation and turn off the power supply (after setting values
to 0) for your fellow students in other sections.
8
Results and Calculations
Part 1: Magnetic Field at the Center of a Coil
For magnetic field at the center of a coil, we can use the following formula:
? =
µ
0
𝑁?
2𝑅
Where
●
B = magnetic field in Tesla (T)
●
I = Current in Amperes (A)
●
R = Radius of coil in meters (m) = 10.5 cm = 0.105 m
●
N = Number of turns in coil = 200
For the first case,
I = 0.4 A
B = -0.01227 mT = -0.00001227 T
Therefore,
µ
0
=
2?𝑅
𝑁?
=
2 • −0.00001227 • 0.105
200 • 0.4
= − 3. 22 • 10
−8
?/?
We can do the above calculation for all the other readings, and hence the experimental value of
can be found. These values can then be used to determine the relationship between the
µ
0
magnetic field B and current I.
Below are the values of the magnetic field we measured for the different values of current:
Current (I) [A]
B_z (mT)
Experimental value of
µ
0
(H/m)
0.4
-0.01227
-3.22
10
-8
•
0.6
-0.001857
-3.25
10
-9
•
0.8
-0.005424
-7.12
10
-9
•
1
-0.005498
-5.77
10
-9
•
1.2
-0.009436
-8.26
10
-9
•
1.4
-0.007141
-5.36
10
-9
•
1.6
-0.006081
-3.99
10
-9
•
1.8
-0.01936
-1.13
10
-8
•
2
-0.006871
-3.61
10
-9
•
Below is the graph obtained for the measurements attained:
9
Part 2: Magnetic Field Along the Central Axis of a Single Coil
Since
, where:
? =
µ
0
𝑁?𝑅
2
2
(𝑅
2
+𝑧
2
)
3
●
I = current in the coil = 2A
●
N = Number of coils = 200
●
R = Radius of coil = 10.5cm = .105m
●
B = Magnetic field associated with the coil (mT)
●
z = distance from the coil along the central axis
For the first case:
I = 2A
B =
0.4636
± 0.04272
µ
0
=
2?
(𝑅
2
+𝑧
2
)
3
𝑁?𝑅
2
=
2 • 0.4636 •
(0.105
2
+0
2
)
3
200 • 2 • 0.105
2
=
2. 43 • 10
−2
?/?
Uncertainty calculation:
∆µ
0
µ
0
=
∆?
?
⇒ ∆µ
0
= µ
0
∆?
?
=
(2. 43 • 10
−2
) •
0.04272
0.4636
=
2. 24 • 10
−3
?/?
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10
We can do the above calculation for all the other readings, and hence the experimental value of
can be found. These values can then be used to determine the relationship between the
µ
0
magnetic field B and the distance from the coil along the central axis.
Below are the values of the magnetic field we measured for the different values of distance from
the coil along the central axis:
z position (cm)
B_z (mT)
Experimental Value of
µ
0
(H/m)
0
0.4636 ± 0.04272
2.43 • 10
-2
± 2.24 • 10
-3
2
0.6115 ± 0.05193
3.39 • 10
-2
± 2.88 • 10
-3
4
0.8029 ± 0.0552
5.17 • 10
-2
± 3.55 • 10
-3
6
1.076 ± 0.05636
8.63 • 10
-2
± 4.52 • 10
-3
8
1.473 ± 0.06006
1.54 • 10
-1
± 6.27 • 10
-3
10
1.874 ± 0.06282
2.59 • 10
-1
± 8.69 • 10
-3
12
2.226 ± 0.06261
4.09 • 10
-1
± 1.15 • 10
-2
14
2.5 ± 0.06898
6.08 • 10
-1
± 1.68 • 10
-2
16
2.48 ± 0.05868
7.88 • 10
-1
± 1.87 • 10
-2
18
2.167 ± 0.06146
8.89 • 10
-1
± 2.52 • 10
-2
20
1.682 ± 0.05851
8.79 • 10
-1
± 3.06 • 10
-2
22
1.251 ± 1.318
8.22 • 10
-1
± 8.66 • 10
-2
24
0.9423 ± 0.05921
7.68 • 10
-1
± 4.83 • 10
-2
26
0.7715 ± 0.04806
7.71 • 10
-1
± 4.81 • 10
-2
28
0.6608 ± 0.0524
8.01 • 10
-1
± 6.35 • 10
-2
30
0.5348 ± 0.0474
7.79 • 10
-1
± 6.90 • 10
-1
Below is the graph obtained for the measurements attained:
11
Part 3: Magnetic Field Along the Central Axis of a Helmholtz Coil
We use the following equation to calculate the magnetic field of a Helmhotlz coil along its
central axis:
, where:
? =
µ
0
𝑁?𝑅
2
2
(𝑅
2
+𝑧
2
)
3
●
I = current in the coil = 2A
●
N = Number of coils = 200
●
R = Radius of coil = 10.5cm = .105m
●
B = Magnetic field associated with the coil (mT)
●
z = distance from the coil along the central axis
For the first case:
I = 2A
B =
0.3335
± 0.03949
µ
0
=
2?
(𝑅
2
+𝑧
2
)
3
𝑁?𝑅
2
=
2 • 0.3335 •
(0.105
2
+0
2
)
3
200 • 2 • 0.105
2
=
1. 75 • 10
−2
?/?
Uncertainty calculation:
∆µ
0
µ
0
=
∆?
?
⇒ ∆µ
0
= µ
0
∆?
?
=
(1. 75 • 10
−2
) •
0.03949
0.3335
=
2. 07 • 10
−3
?/?
We can do the above calculation for all the other readings, and hence the experimental value of
can be found. These values can then be used to determine the relationship between the
µ
0
magnetic field B and the distance from the Helmholtz coil along the central axis.
Below are the values of the magnetic field we measured for the different values of distance from
the Helmholtz coil along the central axis:
z (cm)
B (mT)
Experimental Value of
µ
0
(H/m)
0 0.3335±0.03949
1.75 • 10
-2
± 2.07 • 10
-3
2 0.4171±0.0417
2.31 • 10
-2
± 2.31 • 10
-3
4 0.5655±0.05282
3.64 • 10
-2
± 3.40 • 10
-3
6 0.6778±0.05455
5.44 • 10
-2
± 4.38 • 10
-3
8 0.9007±0.05619
9.40 • 10
-2
± 5.86 • 10
-3
10 0.8997±0.05676
1.24 • 10
-1
± 7.85 • 10
-3
12 1.225±0.05878
2.25 • 10
-1
± 1.08 • 10
-2
14 1.23±0.06378
2.99 • 10
-1
± 1.55 • 10
-2
12
16 1.589±0.6353
5.05 • 10
-1
± 2.02 • 10
-1
18 1.879±0.06647
7.71 • 10
-1
± 2.73 • 10
-2
20 1.804±0.06143
9.43 • 10
-1
± 3.21 • 10
-2
22 1.444±0.0587
9.49 • 10
-1
± 3.86 • 10
-2
24 1.456±0.05978
1.19 ± 4.87 • 10
-2
26 2.047±0.06467
2.05 ± 6.47 • 10
-2
28 1.522±0.05976
1.85 ± 7.25 • 10
-2
30 1.279±0.06226
1.86 ± 9.07 • 10
-2
32 0.9796±0.05484
1.70 ± 9.50 • 10
-2
34 0.7349±0.057257
1.50 ± 1.17 • 10
-1
36 0.5449±0.479
1.30 ± 1.15
38 0.3543±0.03801
9.85 • 10
-1
± 1.06 • 10
-1
Below is the graph obtained for the measurements attained:
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13
Discussion and Conclusions
In regards to our calculations, first we calculated the magnetic field at the center of the
coil by using
“
to identify the experimental value of
,
which was -3.22*10^-8 H/m.
? =
µ
0
𝑁?
2𝑅
"
µ
0
Once this was found, we could apply the formula to the other readings we obtained and come to
an understanding of the relationship between
Magnetic field B
and
Current I.
The formula was
easy to use since all the values of the unknowns were either given or obtained from the
experiment so all that was needed to be done was for it to be plugged in. Since we started from
0.4 amps up t0 2 amps, a total of nine trials were conducted not including the mess ups and with
all the pieces of data, a ‘
Magnetic field vs Current’
graph was constructed. From the graph, a
general conclusion stating that, as the current increases, the magnetic field increases
proportionally can be stated.
Furthermore, after performing our second part of the experiment, we calculated the
magnetic field along the central axis of a single coil by using a similar but different formula
known as “
” to once again obtain the value of
.
By using the information gained
? =
µ
0
𝑁?𝑅
2
2
(𝑅
2
+𝑧
2
)
3
µ
0
from the experiment and data already given we were able to plug in the values, solve and get
2.43*10^-2 H/m. For this experiment we executed a total of 16 trials in which this formula could
be applied and help us determine the relationship between magnetic field B and the distance
from the coil along the central axis. With both the values of the magnetic field and its positions
established, a
Magnetic field vs Position
was created and conclusions drawn from it told us that
the magnetic field is inversely proportional to the position, so as position increases the magnetic
field decreases and as it increases the field decreases causing the graph to look like a bell curve.
Finally, the last part of the experiment was done where once again we found the value of
using the formula
” which occurred to be 1.75*10^-2 H/m. We implemented
µ
0
"? =
µ
0
𝑁?𝑅
2
2
(𝑅
2
+𝑧
2
)
3
14
this calculation on some, but not all thirty trials to help us determine the relationship between
the magnetic field B along the central axis of the Helmholtz coil. Once the final graph of this
experiment was created, it was in the shape of a quartic function but we can observe how it stays
almost the same until the middle. The conclusion drawn from the graph was that the magnetic
field along the central axis of the helmholtz coil produces a uniform magnetic field.
To conclude, by experimenting with three different motives surrounding magnetic fields,
we were able to understand the concepts of how magnetism has an influence on moving charges
and how different factors surrounding coils or the strength and direction of magnetic fields
influence different magnetic properties.
15
References
Kraft, Franz. “Helmholtz coil.”
Wikipedia
, https://en.wikipedia.org/wiki/Helmholtz_coil.
Accessed 27 January 2023.
“What are magnetic fields? (article).”
Khan Academy
,
https://www.khanacademy.org/science/physics/magnetic-forces-and-magnetic-fields/
magnetic-field-current-carrying-wire/a/what-are-magnetic-fields. Accessed 27 January
2023.
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- Q1. a) An SCR based AC power controller circuit is given in Figure 1. The circuit connects a 240V 50Hz AC source vi to a resistive load with resistance of 100. If the delay angle is a-15°, draw the waveforms of the load voltage Vo, and the voltage across SCR2. Calculate the power supplied to the load. + Vi Va + 2 SCR2 H₁₁ SCR₁ Figure 1 SCR based AC power control circuit Va + 2 io b) Figure 2 is a single phase 2-level voltage source converter (VSC) with a DC voltage Vd-200V. A PWM scheme, by comparing a reference value with a 10kHz triangle waveform, is used to control the switches S₁ and S2. The modulation index of the PWM is m=0.8. With the aid of graphs explain how pulse signals for the switches are generated, sketch the output voltage Vo, determine the duty cycle, and calculate the average output voltage. + Vo + Vo √ ₂ == Figure 2 A single phase VSCarrow_forward3.21 я A three-phase half-wave controlled rectifier is connected to a star supply and a resistive load R =10 2. If the rectifier is turned-on at a firing angle of 60°, calculate average and rms output voltages and currents. Assume phase supply voltage to be 120 V. [Ans: 70 V, 106.8 V, 7 A, 10.68 A]arrow_forward! Required information Problem 03.093- The dc output voltage in the half wave rectifier The half-wave rectifier in the figure given below is operating at a frequency of 60 Hz, and the rms value of the transformer output voltage is 19.6 V ± 10%. Assume the value of R to be infinite. ли D₁ CR Problem 03.093.a - The dc output voltage in the half wave rectifier Calculate the nominal value of the dc output voltage Vo if the diode voltage drop is 1 V. (Round the final answer to one decimal place and Include a minus sign if necessary) The nominal value of the dc output voltage Vois 29.5 V AURA T 5 + VO Q Search 6 P & 7 A S Po JA Rarrow_forward
- 1. What is the average output voltage of a three phase three pulse rectifier operating on a line to linevoltage ‘X’ V? (Here X=1813266) 2. What is the average output voltage of a three phase six pulse rectifier operating on a line to neutralvoltage ‘Y’ V? (Here Y= 181326) 3. What are the advantages of a three phase six pulse rectifier over a three phase three pulse rectifier?arrow_forwardA Half-Wave rectifier with a capacitor filter (capacitance is 4700 ?? ) supplies to a load an average current of 1.4 A at an average voltage of 19.5 V. Determine: a) the ripple factor, b) ????.arrow_forwardA half-wave rectifier is needed to supply 15 VDC to a load that draws an average current of 250 mA. The peak-to-peak ripple is required to be 0.2 V or less. a) What is the minimum value allowed for the smoothing capacitance? b) What if a full-wave rectifier is needed?arrow_forward
- The figure to the right illustrates the market for coffee in Colombia. Suppose the demand for coffee in Colombia is DColombia and the supply in Colombia is Scolombia: The world price of coffee with free trade is Pw 1.) Using the point drawing tool, indicate the quantity of coffee demanded at the world price (Q) in Colombia and the quantity of coffee supplied at the world price (QS) in Colombia. Properly label each point 2.) Using the point drawing tool, indicate the domestic market equilibrium price and quantity for coffee in Colombia if trade were completely eliminated. Label the point 'E' Carefully follow the instructions above, and only draw the required objects. PW Price per pound Quantity (Pounds of Coffee) SColombia Peolombiaarrow_forwardA P KN Refer to the figure. d₁ 0.9 m Force P is 60 kN and each force Q is 125 kN. Q KN B Add your answer Integer, decimal, or E notation allowed Q KN The diameter d₁ = 30 mm and diameter d₂ = 50 mm. d₂ 1.2 m What is the normal stress in rod BC in MPa? Round to the nearest integer and enter the magnitude only. Carrow_forwardDerive the equation of the Vdc, Vrms, Efficiency, RF, and TUF of a given three-phase half-wave rectifier circuit shown. Draw the output waveform of the pulsating DC voltage neatly and accurately.arrow_forward
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