electromagnetic induction

roshanhero

Please help me to solve question no.6(b) and no.6c(ii) of the attached question.Or just give me the hint.

9702_w05_qp_4.pdf145.1KB

Reply 1

**mentalheroin**

roshanhero

Please help me to solve question no.6(b) and no.6c(ii) of the attached question.Or just give me the hint.

For 6b.............. the magnetic flux is equal to the area multiplied by the magnetic flux density (magnetic flux = BA, A is area, B is magnetic flux density)

Then, for the magnetic flux linkage, you multiply this value by N(no of turns)......(magnetic flux linkage=BAN)

Reply 2

Drummy

6cii

EMF = \frac{d \phi }{dt}

which in this context is equal to the slope of a graph of BA aginst time.

If the flux is constant there is no EMF.

In the second and third parts there will be a constant EMF but in opposite directions.

Reply 3

Morbo

Drummy

EMF = \frac{d \theta }{dt}

Or more usefully for this question:

EMF = -A\frac{d B}{dt}

Hence, when

\frac{dB}{dt}

(the gradient of the B-t graph) is non-zero, there is an EMF induced proportional to the gradient. When the gradient is a positive, there is a negative voltage induced and vice versa.