Search entries or author Unread     Subscribe 6 e/m help topic All Sections Questions about the lab this week? Ask here.  Reply (https:// Inga Hwang (https://ccsf.instructure.com/courses/41070/users/177478)  Tuesday I'm not sure if I'm just misunderstanding how to use the simulation (or the directions) but it seems to be asking me to enter a value of r in cm, and I thought that's what we were supposed to be measuring for (as in we set the value of V and I and then it outputs a value of r)? Or are we supposed to be coming up with our own value of r? It also has sliding controls for the electric and magnetic fields, are we supposed to be doing anything with either of those? Thank you!  Reply (http Inga Hwang (https://ccsf.instructure.com/courses/41070/users/177478) Tuesday  Also, for part 1 I know the field is supposed to be constant, but I don't understand how we Search know entrieswhat or author Unreadbe? There's the value of B should control to set a value ofB,Subscribe but we  a sliding  aren't given a value of B to set it to, are we just supposed to arbitrarily choose one?  Reply Inga Hwang (https://ccsf.instructure.com/courses/41070/users/177478) (http  Tuesday Actually I just realized that we know the value of B because of the relationship B = 7.48E-4 * I, but are we supposed to adjust the sliding value of B on the simulation to the proper value of B or is that something that would automatically happen?  Reply (http Micheal Bollinger (https://ccsf.instructure.com/courses/41070/users/767)  Tuesday For part 1, adjust the sliding bar to 1.50 A for the B field, then cycle through voltage values For part 2, adjust the sliding bar to 100 V for accelerating voltage, then cycle through current values.  Reply (http Micheal Bollinger (https://ccsf.instructure.com/courses/41070/users/767) Tuesday Ok first step is to setup up your initial conditions for current and go through voltages. There's a little slider bar on the left side to adjust voltage and current.  Search entries or author Unread    Subscribe Then to measure click on the "Measure circle on/off" and play with the slider to get where you think is good, you may need to make an executive decision on the exact value if it's close but not quite there (aka guestimate). Search entries or author Unread    Subscribe Then at the bottom where you can enter stuff, enter your values, and radius and click "check" at the bottom, if it comes up red, it's not the right radius for that V and I, if it's green then it's "good enough"  Reply (http Inga Hwang (https://ccsf.instructure.com/courses/41070/users/177478) Yesterday got it, thanks so much!  Reply  Magnetic Fields – Solutions So the data that was given was the first two columns, the third is just all the math done at once. The magnetic force column is gotten by taking the difference between the reading and the first reading, dividing that by 1000 to convert grams to kg, and then multiplying by 9.8 the acceleration due to gravity to get force. We can use the acceleration due to gravity because that’s what the scale thinks it is feeling a force from. Example: 3rd column row 3, which in this case is actually the 4th row since in my spreadsheet the titles for each column actually start on row 2. But the equation is ‘=(B4-B$3)*9.8/1000’ the ‘$’ in front of the ‘3’ tells the spreadsheet that you do not want to change that row when copying the equation. = (160.661 − 160.645) × B4 B$3 9.8 = 0.0001568 𝑁 1000 0.0045 Current (Amps) 0.004 Magnetic Force (N) 0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 -0.0005 0 1 2 3 4 5 6 0 0.39 1.03 1.48 1.95 2.37 3.09 3.6 4.03 4.43 5.27 Scale Reading (g) 160.645 160.661 160.727 160.76 160.795 160.837 160.891 160.927 160.961 161 161.046 Magnetic Force (N) 0 0.0001568 0.0008036 0.001127 0.00147 0.0018816 0.0024108 0.0027636 0.0030968 0.003479 0.0039298 Current (A) Here’s the results of the LINEST function, all I really care about is the slope, the y intercept (b) should ideally be zero as there should be no magnetic force when there is no current, however real world data taking usually doesn’t make things go as “ideally”, that said the fact the uncertainty of that is larger than the value definitely does allow 0 to fit within the error so that is a good thing. value unc. m B 0.000772987 -2.2395E-05 1.29408E-05 3.88164E-05 Now the slope in this case is going to be the magnetic force over the current, slope is rise (y-values) over run (x-values), and we know that the magnetic force can be represented as F=ILB so, slope = LB. L in this case is 1.00 ± 0.05 cm so to get B we divide slope (LB) by L and be careful to convert to SI units! 𝐵= 0.000773 = 0.0773 𝑇 0.01 The uncertainty we use the division rule for uncertainty or take the relative uncertainty of values that we use in our equation in this case m & L 𝛿𝐵 = 𝐵 � 0.05 1.29 × 10−5 𝛿𝐿 𝛿𝑚 + � = 0.0773 � + � = 0.005 𝑇 0.000773 𝑚 1 𝐿 Some may be confused why I didn’t convert units on the length here, this is because I’m taking the relative uncertainty which always has the same units in numerator and denominator so those would cancel out to a unitless quantity and also the conversion factor would cancel out as well, so numerically it’s the same. But more importantly it’s just old habit of mine to do it this way so that there are less possible areas to mathematically goof up by doing an improper conversion or something. 𝐵 = 0.078 ± 0.005 𝑇 For part 2 it’s a similar setup 0.006 L (cm) Magnetic Force (N) 0.005 1 2 3 4 6 8 0.004 0.003 0.002 Scale Magnetic Reading (g) Force (N) 160.735 0.000882 160.822 0.0017346 160.849 0.0019992 160.919 0.0026852 161.06 0.004067 161.165 0.005096 0.001 0 0 2 4 6 8 10 Wire Length (cm) value unc. m b 0.000600106 0.000343576 2.45593E-05 0.000114317 Here I could have converted the length into meters before doing the LINEST function, however I decided to do it afterwards, it makes no difference however other than my values will differ from yours by a factor of 100 until I calculate the value of B. Now here for the y-intercept is a case where the uncertainty does not encompass the value 0 N at a length of 0 cm, which is a red flag that some of the data may have been taken a bit sloppily… but since I’m the person who took the data and I’m the teacher I’m going to let it slide  plus I already gave you the assignment I’m not about to redo all the data and tell you to do it over  But these are the things to keep an eye out for when you are taking the data and looking at the results. Now in this case the slope is force over length and again F=ILB so slope = IB, where I was 1.00 ±0.04 A, not forgetting my length is in cm for this, so multiply by 100 for the conversion, if you already converted before finding the slope then you’d exclude that step 𝐵= 0.0006 𝑁 100 𝑐𝑚 × = 0.06 𝑇 1.00 𝐴 ∙ 𝑐𝑚 1𝑚 𝛿𝐼 𝛿𝑚 0.04 2 × 10−5 𝛿𝐵 = 𝐵 � + � = 0.06 � + � = 0.004 𝑇 0.0006 𝐼 𝑚 1 𝐵 = 0.060 ± 0.004 𝑇 Discrepancy Check 𝐷𝑖𝑠𝑐𝑟𝑒𝑝𝑎𝑛𝑐𝑦 = |𝐵1 − 𝐵2 | = 0.0773 − 0.06 = 0.0173 𝑇 𝑇𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 = 2 × (𝛿𝐵1 + 𝛿𝐵2 ) = 2(0.005 + 0.004) = 2 × .009 = 0.018 𝑇 So just barely passed the discrepancy test with a larger tolerance, I suspect that bit of “sloppiness” that was seen earlier for the second part y-intercept is why it was so close. Special note If I didn’t give you the weight of the magnet you could have done the lab just as easily, except instead of your y-axis being “Magnetic force” it would have simply been “Combined force”, the slope would have been identical and since that’s all you need to calculate the magnetic field you’re set! That said why didn’t I do it this way? Because I wanted you to make the relationship between F=ILB and slope, where F in this case is the magnetic force. Otherwise you’d have something like F_combined = F_magnetic + F_grav = ILB + mg which isn’t quite as straight forward, but do note that if I didn’t give you the weight of the magnetic you could have figured it out from linear regression, just use the part we didn’t need, i.e. y-intercept. Magnetic Force discussion 1. First the wrong answer would be “the vertical piece are not in the magnetic field” as the magnetic field extends beyond that just the small area between the poles, besides there is some vertical parts between the magnetic poles of each magnet. There are two answers to this technically. a. First being that the direction of the force would be perpendicular to the direction of the current and magnetic field which would mean the force would be sideways so wouldn’t create a downward or upward force on the scale hence not change the scale reading b. Second because of the symmetrical nature of the vertical pieces and the fact the current travels in opposite directions on each side means that the forces would balance out, so even though the force is sideways if one is pushing outward the other is also pushing outward in the other direction, so there’s 0 net force contribution when you take into account both sides in any direction. 2. The scale measures the force that is pushing down on it, it doesn’t care what is creating that force but it assumes that the force of gravity is doing it as a result shows the resultant mass that corresponds to the force that is pushing down on it. So the force of gravity isn’t changing meaning that the difference in mass reading between no current and whatever current Is turned on would be due to the magnetic force pushing down on the magnet. 3. If the scale reading goes down that means there’s less force pushing on the scale, so the force on the magnet is UP, however Newton’s 3rd law states opposite & equal, so the force on the magnet being up corresponds to a force on the wire being down. F vs 1.579 y = 0.0008x + 1.5743 R2 = 0.9975 1.578 1.577 2 1.576 1.575 1.574 0 1 2 3 3 5 5 LD 1(A) 0.0045 0.004 Current (Amps) 0 0.0035 0.003 0.0025 Magnetic Force (N) 0.002 0.39 1.03 1.48 1.95 2.37 3.09 3.6 Scale Magnetic Reading (8) Force (N) 160.645 0 160.661 0.0001568 160.727 0.0008036 160.76 0.001127 160.795 0.00147 160.837 0.0018816 160.891 0.0024108 160.927 0.0027636 160.961 0.0030968 0.003479 161.046 0.0039298 0.0015 0.001 0.0005 0 4.03 4.43 5.27 161 -0.0005 5 Current (A) Here's the results of the LINEST function, all I really care about is the slope, the y intercept (b) should ideally be zero as there should be no magnetic force when there is no current, however real world data taking usually doesn't make things go as "ideally”, that said the fact the uncertainty of that is larger than the value definitely does allow to fit within the error so that is a good thing. m B value 0.000772987 1.29408E-05 -2.2395E-05 3.88164E-05 unc А Current B с Magnetic Force scale reading 0 0 0.39 0.0001568 1.03 0.0008036 1.48 0.001127 1.95 0.00147 2.37 0.0018816 3.09 0.0024108 3.6 0.0027636 0.0030968 4.43 0.003479 5.27 0.0039298 160.645 160.661 160.727 160.76 160.795 160.837 160.891 160.927 160.961 161 161.046 slope y-intercept 0.000772987 -2.23953E-05 AutoSave OFF HESH tutor_1 Home Insert Draw Page Layout Formulas Data Review Share Comments х = Times New Roman V 12 - Å Α' Α' View Tell me % Conditional Formatting Format as Table Number Cell Styles V 02 三三三國· Paste BI U a. A v A V Editing Cells Analyze Data D18 4 x ✓ fx B с D E F G H 1 J к L M N 0 P Q 1 2 3 V vs r2 4 Voltage (V 100 125 1-1.5A B-1.12E-3T 5 350 6 150 300 7 Radius (m) 0.031 0.034 0.037 0.04 0.042 0.045 0.047 0.049 0.051 Radius (m) 0.000961 0.001156 0.001369 0.0016 0.001764 0.002025 0.002209 0.002401 0.002601 175 200 225 . 8 250 200 9 10 11 12 250 275 300 Voltage (V) 150 13 100 14 15 SO 16 17 slope 120858.7364 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 18 19 Radius? (m) 20 21 22 23 24 25 26 e to m for a fixed current e to m for a fixed voltage + Ready B + 100%
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