Q21(c) 2009 HSC

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Mini
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Joined: Sat Mar 06, 2010 10:33 am

Q21(c) 2009 HSC

Postby Mini » Wed Jul 07, 2010 11:03 am

Hello Prof Joe,

I've been wondering about this question for a while... Here's the paper: http://www.boardofstudies.nsw.edu.au/hs ... hysics.pdf

There are two approaches to what you could take.

1) Use F=BIL and then use F=mg and solve for B.

2) Halve the torque found in part (b) and use it in T=nBIA and then solve for B.

I know why the 1st approach works but I'm not sure why the 2nd approach would work. Why does it work?

Thanks

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joe
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Location: Sydney

Re: Q21(c) 2009 HSC

Postby joe » Wed Sep 15, 2010 4:58 pm

In that question, you are expected to assume that the magnetic field on the RH side of the coil is negligible.

The formula you give (T=nBIA) is the torque exerted by a coil all of whose area is in field B.

Here, one side is in B, another is assumed to be in zero field, and the other two sides exert no torque anyway because of their orientation. So yes, this coil gives a torque of T = nBIRL = nBIA/2 where R is the radius of the coil and L its width.

Joe


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