Suggested Solutions (Including Photonics as Detailed Study)

[whyareyoureadingthis]

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?

[m@tty]

=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.

I ended up labeling the first emf as qV, where q is an arbitrary value(as they didn't actually give any values for flux) and the final section as -q/2 V.

Because they didn't give any indication of the voltmeters connection. So even though you know which direction the emf will be in you don't know what will come up on the meter.

[SixWinged]

=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.

I wrote a note saying that I took X to Y (I drew it on the diagram) as positive.

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?

Lenz's law IS the negative sign in Faraday's equation. Lenz's law is about the polarity rather than the magnitude.

[kyzoo]

=/ I was thinking about whehter you had to put the first section (t=0 to t=1) above or underneath the x-axis. I came to the conclusion that ti doesn't matter. You don't have a power source with a positive/ngative terminal in the diagram, therefore you can't tell which way si positive current and which way is negative.

I ended up labeling the first emf as qV, where q is an arbitrary value(as they didn't actually give any values for flux) and the final section as -q/2 V.

xD xD I just put 2k and -k

[DuBistKomisch]

I think it may be Lenz's law rather than faradays. Faraday's would not imply any direction, Lenz's law takes into account the fact that voltage is above and below the axis.

I said Lenz's. I was under the impression that Faraday's was emf=d(phi)/dt, whereas Lenz's was the same except with the addition of a negative sign (emf=-d(phi)/dt). Can someone please clarify with an explanation?

Faraday’s law of induction states that the magnitude of the induced emf is equal to the time derivative of flux. When direction is taken into consideration, the sign of the emf is negative:
E=-(∆Φ_B)/∆t
Lenz’s law states that the direction of the induced current is such that the magnetic field produced will be in the opposite direction to the change in flux that induced it.

[DuBistKomisch]

Also, http://en.wikipedia.org/wiki/Lenz's_law says:

Lenz's law is a common way of understanding how electromagnetic circuits must always obey Newton's third law. Lenz's law is named after Heinrich Lenz, and it says:

"An induced current is always in such a direction as to oppose the motion or change causing it"

Faraday's law of induction indicates that the induced electromotive force (emf) and the change in flux have opposite signs, and it also gives the direction of the induced (emf) and current resulting from electromagnetic induction.

[bomb]

for 10 I pretty sure voltage was -3V (negative)

Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...

Doesn't EMF have a modulus sign somewhere that makes it positive?

[Chavi]

for 10 I pretty sure voltage was -3V (negative)

Hmm, good point. I'm not sure if the sign should really matter though since we're not told the polarity of the voltmeter. In fact I don't remember being told whether it is measured at the voltmeter at all...

Doesn't EMF have a modulus sign somewhere that makes it positive?

EMF is a vector quantity - if they were to ask for the magnitude of voltage, then ye, you would mod it

[jackchan1993]

PEOPLE! did u have to label the voltage values? :O

dammit... i didn't do it, i just drew one voltage larger than the other ><

[2iraqi4u]

Hey I heard someone really awesome sat afront of you in the exam Mr. DBK  :smitten:

[DuBistKomisch]

Hey I heard someone really awesome sat afront of you in the exam Mr. DBK  :smitten:

Haha, but he was too Iraqi for me and kept looking at my multichoice answers after the exam.

[Chavi]

PEOPLE! did u have to label the voltage values? :O

dammit... i didn't do it, i just drew one voltage larger than the other ><

There was no way to label voltage. They didn't give figures for the Area of the coil or field. . .

[jackchan1993]

PEOPLE! did u have to label the voltage values? :O

dammit... i didn't do it, i just drew one voltage larger than the other ><

There was no way to label voltage. They didn't give figures for the Area of the coil or field. . .

phewww lols

[m@tty]

Haha, I just gave arbitrary units to make it perfectly clear that the second emf = -0.5 the first. kyzoo did this too. But you didn't need it.

[2iraqi4u]

Hey I heard someone really awesome sat afront of you in the exam Mr. DBK  :smitten:

Haha, but he was too Iraqi for me and kept looking at my multichoice answers after the exam.

Ahaha Aye I was just checking to see if I got it all right! And i think I did woo!