2006 HSC Question (1 Viewer)

Uzername · Nov 2, 2011

I don't get 2006 HSC Question 20 d)... The Excel solutions say the answer is 0.0225m, while these solutions state the answer is 0.23m. In either case, I still don't understand how to reach the answer. Could someone explain?

Exam is here: http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2006exams/pdf_doc/physics_06.pdf

Uzername · Nov 3, 2011

Bump... :|

CalSandercock · Nov 3, 2011

Maybe try using two different values of the the bottom current in the parallel conductors formula and then solve the two equations simultaneously?
Or figure out how the change in weight equals force and then you just sub into the formula

CalSandercock · Nov 3, 2011

Actually I just looked at my success one and the answer is completely different - simultaneous wouldn't have worked because the two distances and two force values are different. here is the success one Answer.

The force between two current-carrying rods is given by Ampere's law ie:
$\frac{F_{b}}{\Delta l}= \frac{k_{m}I_{1}I_{2}}{d}$

Now ${\Delta l}$ = 2.6m, $I_{2}$ = 50A, and $k_{m} = 2 \times 10^{-7} NA^{-2}$

Slope of graph = $\frac{\Delta m}{\Delta I_{1}}$ , where -mg = $F_{B}$ , dividing by $I_{1}$ we have $\frac{-mg}{I_{1}} = \frac{F_{B}}{I_{1}} = -slope \times g.$

Using this data $d = \frac{2 \times 10^{-7} \times 50 \times 2.6}{-slope \times 9.8}.$ Now, the slope from the graph = $\frac{rise}{run} = \frac{0.5464 - 0.5489}{21.2 - 0} = -1.179 \times 10^{-4}kg..A^{-1}$ . Therefore $d = \frac{2 \times 10^{-7} \times 50 \times 2.6}{-(-1.179 \times 10^{-4}) \times 9.8} = 0.0225m$ (which is 0.23m rounded up btw

)

The current carrying rods are 22.5 mm apart

Uzername · Nov 3, 2011

Thanks for the response,
so then what was wrong with the Biki solutions? I have the Success One answers myself but don't get it... Lol

hyhy · Feb 10, 2013

Uzername said:
Thanks for the response,
so then what was wrong with the Biki solutions? I have the Success One answers myself but don't get it... Lol

I was just doing this question the other day so thought I'd clarify this question even though its an old thread. The biki solution still shows the incorrect answer (which confused me as well at first when I was going through the answers)

Biki solution subtracted the mass of the rod from the reading and assumed that this value is the force due to the magnetic field. However you must multiply mass by the acceleration due to gravity (9.8) to correctly convert to weight force, which in this case is equal to the magnetic force.

By not multiplying by 9.8, they end up being off by a factor of 10.

2006 HSC Question (1 Viewer)

Uzername

B.S., M.S., M.A., Ph.D.

Uzername

B.S., M.S., M.A., Ph.D.

CalSandercock

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CalSandercock

New Member

Uzername

B.S., M.S., M.A., Ph.D.

hyhy

New Member

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