Distance between the rods??? It changes! (1 Viewer)

dimivat

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For the question that asks the distance between the rod, if the mass they gives you decreases with the increase in current doesn't that mean the distance is changing as a result of the second rod being atracted to the first?
 

dimivat

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ye but then u can't because the current is 0 in the formular doesn't work, hate questions when they don't ask them right.

I recon it was a fare exam through was a well writen one over all!
 

Sober

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The distance between the rods will not change until the current is high enough to make the reading on the scales equal 0kg, at which point it will rise instantly to stick to the other rod. So no, the distance does not slowly decrease.
 

markus123456789

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Im pretty sure that there was a simpler way to do it. From what i remember i let

Fw=mg = F/l= KI1I2/D

From there the equation could be rearranged to give D as the subject.

D = KI1I2L / mg
I cant remember exactly how it went but from there looking at the table that was given ,take a value for mg (weight force) and the corresponding I2 . From what I remember it had an answer in the …X10-3 Meh Im not sure if I did it right, but oh well. :D Physics is over
 

rampeh

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The force isn't enough to counterract gravity' acceleration, so the rods stay stationary, the mass and weight force on the scale would be 0 if it lifted off the scale, and there was no indication the scale was depressed in the question so we can only assume the distance was constant.
 

dimivat

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it must change becuase there is a force between them to attract one another only small amounts but it does change. Y else would the weight change?
 

Sober

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dimivat said:
it must change becuase there is a force between them to attract one another only small amounts but it does change. Y else would the weight change?
Stand on the scales at your house, now push down on a table next to you with your hand, the reading can go down without you lifting off the scales. Sheesh, you do physics?
 

dimivat

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Mate that is because the scale moves up slightly when you do that trying to returen to equalibrium these rods are fixed.

yes i do physics

Lets just focus on it being over!
 

hyparzero

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Idiots, you know that graph you drew previously? You use the line of best fit, and read off the weight from when Current = zero
 

nittyc

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BTW i think to get the 3 marks u have to use the gradient of the line of best fit to get M/I2, and u sub that in- u cant use any random point because the y intercept is not zero! u will definitely get the wrong answeri got them like 5cm apart or something by that method.
 

yankyfly

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the gradient was irrelevent. You just picked one point on your line of best fit, and examined the current that was flowing and the change of mass that had occured between when that current was flowing and when it was not. You times'd that mass change by 9.8 to get the force that was happening when that current was flowing. then you subeed your values into the equation.

Depending on your line of best fit. The distance apart should have been between 2-3 cm
 

donrob15

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but if you put a current of zero (I1 or 12) into the F/l equation you will get nothing as you are multiplying by zero. you have to figure out the force of attraction. since it is a proportional relationship, all you need to do is pick one of the currents that were recorded (i picked the highest one...), find the difference between the actual mass of the rod (which you get off the graph when current = 0) and the measured mass of the rod, and times it by 9.8. You now have all the the values you need for the F/l equation, and the distance between the two rods does not change, so you just rearange and solve to find d. the distance not changing can be explained if you think of vector fields. The force of attraction is opposing the effect of gravity (that is, wieght) so the recorded 'mass' can actually decrease without the rod actually moving. the rod will not move until the force of attraction matches the weight of the rod, in which case it will lift off the balance and stick to the upper rod. :) hope you all did well.
 

lonely_devil

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use the line of best fit to work out the gradient... n then u can find the distance through that...
since equation of a straight line is y=mx+b
u just rearrange the force between two conductors formula... and u'll get something like this:
F = kI(upper)I(lower)L/d
and since I(lower) is the x values...
F = kI(upper)L/d x I(lower)
so the gradient = kI(upper)L/d
n then just sub everything in...

i hope that's right... =]

as for the mass... i had no idea... i worked out the distance... then subbed it in to get the mass in some strange way... =P
 

nittyc

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this is how i believe you solve part B:
Find the gradient of the lobf (works out to be mass of rod/current (kg/A))
equate f=ma and f=(k*i1*i2*l)/d
so
ma=(k*i1*i2*l)/d
a=(k*i1*i2*l)/d
d=(k*i1*i2*l)/a now use ur line of best fit

d=k*i1*l/gradient*a
bang in the values

u cannot use an artbitary point, because ur ratio will be incorrect. You can only use it when u have a y=ax curve, not when its y=ax+b
 

suzieee

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markus123456789 said:
Im pretty sure that there was a simpler way to do it. From what i remember i let

Fw=mg = F/l= KI1I2/D

From there the equation could be rearranged to give D as the subject.

D = KI1I2L / mg
I cant remember exactly how it went but from there looking at the table that was given ,take a value for mg (weight force) and the corresponding I2 . From what I remember it had an answer in the …X10-3 Meh Im not sure if I did it right, but oh well. :D Physics is over
i copied the formula I1I2 formula from the formula sheet and wrote down what i knew, so i should get one mark shouldnt i?

im happy that physics is over too... now i only have maths ag and chem left... my favourites
 

nittyc

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yankyfly said:
the gradient was irrelevent. You just picked one point on your line of best fit, and examined the current that was flowing and the change of mass that had occured between when that current was flowing and when it was not. You times'd that mass change by 9.8 to get the force that was happening when that current was flowing. then you subeed your values into the equation.

Depending on your line of best fit. The distance apart should have been between 2-3 cm
thats complete bullshit. the gradient is absolutely relevant; thats why they got u to draw the graph.
 

Lions_Fist

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Yeah I got 2.3cm also... but I used g=10 instead of 9.8 for some reason when converting to newtons... Ohh well, I hope they decide not to take marks off for it :p
 

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