HSC Stuff > HSC Physics

Physics Question

(1/1)

bloop:
My trials are coming up soon and I need help with this question:

jakesilove:

--- Quote from: bloop on August 05, 2017, 11:59:43 am ---My trials are coming up soon and I need help with this question:


--- End quote ---

Hey! First, you need to think about whether the force is attractive or repulsive. Since the currents are in opposite direction, both other wires create a repulsive force at conductor C.



Now, we need to calculate the magnetude of the force. We have a very useful equation:



We know that the wire is of unit length, and we know the distance between the wires. So, we know that the force on Conductor C from conductor A is



There are two wires, so the net forces add up to

Younem:
Hi, I had this question too. It's from the 2015 catholic trail paper but the answer booklet said that the answer was B. I did the same as you, should I ignore the answers?

kiwiberry:

--- Quote from: Younem on August 07, 2017, 09:34:41 pm ---Hi, I had this question too. It's from the 2015 catholic trail paper but the answer booklet said that the answer was B. I did the same as you, should I ignore the answers?

--- End quote ---

I think you have to take into account the fact that the two forces act in different directions, so the net force will be less than 3.2 x 10-6 N. But the only way I can work out the exact value is using the sine rule which I don't think is assessable in physics, unless I'm missing something :/

jamonwindeyer:
You can get this done without the sine rule if you are super clever and spot that the force on that conductor will be directed exactly between the two other conductors - This comes from certain components of the force vectors cancelling out.

You won't need to do anything like this in the HSC, but the way to do it is to take the answer and multiply it by \(\cos{30}\).



That's where B comes from ;D the CSSA assessing this is technically allowed but I doubt you'd see it in a HSC paper, and I doubt you'd see it again in CSSA either :P

Edit: Visual explanation --->

Navigation

[0] Message Index