The primary coil of a step-up transformer is connected to a source of alternating pd.
The secondary coil is connected to a lamp.
Which line, A to D, in the table correctly describes the flux linkage and current through
the secondary coil in relation to the primary coil?
secondary magnetic flux linkage ....................secondary current
primary magnetic flux linkage......................... ...primary current
The answer is A and I know why the first column is greater than one, but why is the current ratio less than 1?
The second question is:
A coil of 50 turns has a cross-sectional area of 4.2 × 10–3m2. It is placed at an angle to
a uniform magnetic field of flux density 2.8 × 10–2T, as shown in the diagram, so that
angle ? =50degrees.
What is the change in flux linkage when the coil is rotated anticlockwise until ? =0degrees?
A The flux linkage decreases by 2.1 × 10–3 Wb turns.
B The flux linkage increases by 2.1 × 10–3 Wb turns.
C The flux linkage decreases by 3.8 × 10–3 Wb turns.
D The flux linkage increases by 3.8 × 10–3 Wb turns.
I used BAN cos theta but am not getting the correct answer which is B. Can someone take me through it if possible. Thanks in advance.
Using conservation of energy between the primary and secondary
If V in the secondary is greater than V in the primary then...
It depends which angle is referred to in the diagram.
I suspect the problem is (guessing without the diagram) that the angle given in the question is not the one you use in the formula. The one in the formula is the angle between the magnetic field and a line at right angles to the coil.
The one you use in that case is 90 minus the one given.
Thanks for the help. That's totally cleared up part 1 for me but part 2 I still can't do hehe. I will try phoning my teacher tomorrow and see if she can talk me through it as she'll have the diagram =O)
for question 2.
Initially flux linkage = BANcos50,
and finally flux linkage = BAN cos 0 = BAN x 1.
So the change in flux linkage = final flux linkage - initial flux linkage = (BAN x 1) - (BAN cos 50) = BAN (1-cos50)