Recent content by Steve Spence

Hallelujah ! :smile: Many thanks TSny, that is a brilliant explanation. My mistake had been assuming the net pressure, f2, was zero and/or not appreciating the double minus in f2-f1.

Thanks for your help guys, but I'm still thinking about this one...may take a while !

Hi Kavya, Which of the 2 solutions are you referring to as erroneous, the one using the energy density equation or the one using P=F/A=FLB/A?

ok : $$F=ILBsin\theta = \frac{(5000\ amps)w}{2\pi(5\ cm)}L(10\times 10^{-3}T)sin \ 90$$ The pressure is then $$P=\frac{F}{A}=\frac{F}{wL}=\frac{(5000\ amps)(10\times 10^{-3}\ T)\cancel{wL}}{2\pi(5\ cm)\cancel{wL}}=159\ Pa$$ So the energy density for the averaged field value of 10mT is 79.6...

OK, the magnetic field at the outer wire is 10mT (20+0)/2. The energy density at the wire is now 79.6 Pa, but I want it to be 318 Pa ! The energy density at any radius between the the outer wire and inner wire is never more than 159 Pa, so my idea that energy density = pressure must be inaccurate ?

I would choose 20mT. I am happy with this value for the field just inside the outer wire, and therefore for the pressure on it : 318 Pa. But I thought the energy density (perhaps clumsily) was the same as pressure. Why is that only 318/2=159 Pa ?

Thanks Gordianus, this kind of makes some sense, but sorry I'm still a bit confused. Why don't both methods give the same answer, as they both assume the same value of the field (20mT) ?

Hi, The problem I am working on requires me to work out the the pressure on the outer conductor of a coaxial cable due to the current on the inner one. This cable carries a dc current of 5000 Amps on the inner wire of radius 2 cm. The outer cylindrical wire of radius 5cm carries the return...

I am currently teaching myself introductory physics, by reading a book called "Physics" (Serway and Jewitt, 9th edition), and other support material. I would really appreciate help when I get stuck !