Recent content by LilyY

Hi Bruce, Thanks for your reply. Would you mind having a look at my post just above yours? I'm confused about whether or not V is path independent. Lily :-)

Hi Yukoel, Thanks for your reply- I can now see that the potential difference across the slider is the expression given in the question. However, why wouldn't it also be correct to find V along the U shaped wire? This would give a different expression, i.e. R'-R would replace R in the...

Hi Bruce, thanks so much for your help. So I've found that the new resistance is R+α(2(L+x'cos(ωt))+l), and the EMF is -B'ωlx'sin(ωt). So I=V/R=-B'ωlx'sin(ωt)/(R+α(2(L+x'cos(ωt))+l)). This implies that V=IR=-B'ωlx'sin(ωt), which is the original EMF, not the answer given. Where am I going wrong?

Homework Statement See attachment. Homework Equations The Attempt at a Solution For the first part: (B' and x' are B subscript 0 and x subscript 0 respectively.) \phi=BA=B'xl=B'l(L+x'cos(ωt)) EMF=-d\phi/dt=-B'ωlx'sin(ωt) Therefore V=-EMF=B'ωlx'sin(ωt) So...