Post by: AlanMate on January 29, 2018, 11:44:04 am
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/physics_06.pdf
Q20. (d) ^
Post by: clovvy on January 29, 2018, 09:41:32 pm
Hello I'm having some trouble with this question. I just cant understand the last part of the question. Calculate the distance between the two copper rods.
https://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/physics_06.pdf
Q20. (d) ^
As the current through the lower rod increased, the mass read by the electronic balance decrease. This indicates that the magnetic force resulting from the current in the rods acted to attract the rods together, producing a force upwards on the lower rod and a reduction in the mass read on the balance. To make this magnetic force, the current in both rods must be travelling in the same direction relative to each other.....
I hope this helps
oops my bad I was looking at A,
for d you need to use Ampere's law... and I don't know how to type up all those fancy stuff (I wish I could) like surds or fractions etc... could've made my explanation a lot easier.....
Spoiler
you know that F(B)/#L=k(m)I(1)I(2) (where # is delta)
now #L=2.6m, I(2)=50 A and k(m)= 2 x 10^-7 NA^-2
slope of graph= #m/#I(1), where -mg=F(B), therefore slope x g=F(B)/I(1)
using the known data, d= (2 x10^-7 x 50x 2.6)/ (-slope x 9.8).. (slope of gradient is rise/run)
=(0.5464-0.5489)/21.2-0 = -1.179 x 1o^4 kg.A^-1.
Therefore d= (2 x10^-7 x 50x 2.6)/ -((-1.179-10^-4) x 9.8)=0.00225
The current carrying the rods are 22.5mm apart...
now #L=2.6m, I(2)=50 A and k(m)= 2 x 10^-7 NA^-2
slope of graph= #m/#I(1), where -mg=F(B), therefore slope x g=F(B)/I(1)
using the known data, d= (2 x10^-7 x 50x 2.6)/ (-slope x 9.8).. (slope of gradient is rise/run)
=(0.5464-0.5489)/21.2-0 = -1.179 x 1o^4 kg.A^-1.
Therefore d= (2 x10^-7 x 50x 2.6)/ -((-1.179-10^-4) x 9.8)=0.00225
The current carrying the rods are 22.5mm apart...
Anyway that was messy because I don't know how to type mathematical notations properly but that is how you do d, hope this helps... and for this question you might want to look at the graph carefully....
Mod edit: Merged posts :)
Post by: blasonduo on January 30, 2018, 09:49:00 am
Hey! I kinda gotchu of what u meant, but i was thinking doing on a paper and uploading the image could help! Thanks !!
Hey, What Clovvy said was 100% right!! I shall try to make it slightly easier to follow to help :))
Post by: blasonduo on January 30, 2018, 06:21:08 pm
But I think the answer to it is 2.3x10^-2m
Oh! I forgot to add in the force of gravity! Thanks for the correction!
I have fixed it now