Transformer, flux linkage and emf short question

[thoradicus]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20%289702%29/9702_s05_qp_4.pdf

Homework Statement

6b

Homework Equations

flux linkage = BA
B=F/iL

The Attempt at a Solution

from my understanding, if B=F/iL , if the current is zero, shouldn't B be infinite,hence phi infinite or something? As seen in the 1st graph. Why does the graph follow a sine graph instead of phi(max)sin(wt)?

Also, can anyone explain why the phase difference is 90 deg instead of 180? i thought the EMF curve would just follow a cos curve because of the negative sign of faraday's law, hence 180 phase difference.

 

[aralbrec]

The Attempt at a Solution

from my understanding, if B=F/iL , if the current is zero, shouldn't B be infinite,hence phi infinite or something? As seen in the 1st graph. Why does the graph follow a sine graph instead of phi(max)sin(wt)?

Apply Ampere's law along a path inside the core where constant H is assumed to follow.

Also, can anyone explain why the phase difference is 90 deg instead of 180? i thought the EMF curve would just follow a cos curve because of the negative sign of faraday's law, hence 180 phase difference.

There is a derivative involved. If the flux is represented as a phasor ψe^jwt (w = frequency), then E = -dψ/dt = -jwψe^jwt. The j means there is a 90 degree relationship between the two. You could also see this by taking the derivative of flux assuming it is a sin and you'd end up with a cos, but phasors (and diagram) helps to keep all the relative phases straight.