Q21(c) 2009 HSC

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Q21(c) 2009 HSC

Post by Mini »

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?

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Re: Q21(c) 2009 HSC

Post by joe »

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.

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