TrueTears question thread

[Mao]

it can be used for coils in a generator

in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.

[TrueTears]

it can be used for coils in a generator

in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.

Oh okay thanks, so 'l' would just be the length of the side of the square/rectangle I'm dealing with? But doesn't that just give the emf on that single side of the entire square/rectangle? So does that mean I  have to calculate emf = Blv for each of the sides (that are perpendicular/that have a perpendicular component to the mag field) and add them in order to get the 'total' emf?

[TrueTears]

In a regular DC motor, which of the following would decrease the speed of the rotating coil in the motor. The answer is these 3 "Decrease the number of turns of the rotating coil." "Decrease the area of the rotating coil" and "Decrease the current in the rotating coil"

I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?

[Mao]

it can be used for coils in a generator

in this case, it is applicable to square/rectangular coils. each side is calculated separately, and the direction of the force can be determined by the RH palm rule. Hence the net torque can be found.

Oh okay thanks, so 'l' would just be the length of the side of the square/rectangle I'm dealing with? But doesn't that just give the emf on that single side of the entire square/rectangle? So does that mean I  have to calculate emf = Blv for each of the sides (that are perpendicular/that have a perpendicular component to the mag field) and add them?

Ahh, ooops, I did not read your post properly, I was talking about F=IlB.

For your case, emf = Blv will not apply because you will not given v, since the coil is in rotational motion. (It is possible to derive the tangential speed, but that is not within the scope of VCE Physics).

You will be using to work out an average voltage (for a 90 degrees rotation)
Slightly beyond VCE level, you may be using [see http://vcenotes.com/forum/index.php/topic,4476.0.html]

[Mao]

In a regular DC motor, which of the following would decrease the speed of the rotating coil in the motor. The answer is these 3 "Decrease the number of turns of the rotating coil." "Decrease the area of the rotating coil" and "Decrease the current in the rotating coil"

I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?

The coil moves in rotational motion, hence force is not the only factor. , hence the distance from axle at which the force act is also important.

By reducing the area, you will be reducing either (or both) of length (l) and distance from axle (x). Reducing l reduces force, whereas reducing x directly decreases the torque.

[however, this is not a conventional way of achieving slower speeds]

[TrueTears]

Ahhh, thanks I get it. So then how do you explain that increasing the speed of the rotating coil increases the average emf?

Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formula hence emf is larger?

[TrueTears]

In a regular DC motor, which of the following would decrease the speed of the rotating coil in the motor. The answer is these 3 "Decrease the number of turns of the rotating coil." "Decrease the area of the rotating coil" and "Decrease the current in the rotating coil"

I understand 2 of those 3 but "Decrease the area of the rotating coil" how does that slow the rotating coil down? I mean according to F = nIlB, reducing n or I will reduce the force on the coil hence slow the coil down, but what does area have to do with it?

The coil moves in rotational motion, hence force is not the only factor. , hence the distance from axle at which the force act is also important.

By reducing the area, you will be reducing either (or both) of length (l) and distance from axle (x). Reducing l reduces force, whereas reducing x directly decreases the torque.

[however, this is not a conventional way of achieving slower speeds]

Ah yes yes yes, thanks again Mao, you lifesaver!

[Mao]

Ahhh, thanks I get it. So then how do you explain that increasing the speed of the rotating coil increases the average emf?

Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formula hence emf is larger?

yes

[TrueTears]

Ahhh, thanks I get it. So then how do you explain that increasing the speed of the rotating coil increases the average emf?

Is it just because the change in flux now happens in a shorter time so change in time is smaller in the formula hence emf is larger?

yes

Awesome, thanks for the quick and helpful replies !

[TrueTears]

Some students are studying the emf induced by a magnetic field in a coil of wire. Their apparatus consists of 100 turns of wire in a magnetic field of T. The coil forms a square shape, with the coil vertical to the magnetic field, the flux through the coil is Wb. What is the area of the coil?

Do I have to take in consideration the number of turns here? ie flux = nBA ?

Because book specifically mentions not to times BA by n, but teacher says I should include the n. So my question is when do I include n to work out the flux when do I don't include n?

Thanks!

[Mao]

noo,

you only include the number of turns when dealing with emf, that and

[TrueTears]

noo,

you only include the number of turns when dealing with emf, that and

Aiight, thanks!

[TrueTears]

Ok I came up with a "proof" but I'm not sure if this is correct or not. Could someone please check and let me know if it's valid or not.

My proof is: "If you decrease the wavelength of light being shone through the 2 slits (this is basically just the young's experiment) then the pattern of bright/dark bands created on the plate/wall will get narrower(more tightly packed) than the wavelength it was before."

So looking at the diagram, say A is a fixed point on the wall. Now let L1 be original wavelength and L2 be the decreased wavelength. Let n1 be the integer in front of L1 and n2 be the integer in front of L2. The dotted line is the central maximum which produces a bright band in the middle of the wall. (This is the case for every young's slit experiment anyway).

Now case 1: Assume that both wavelength L1 and L2 produce a point at A which is a bright band (light band) due to constructive interference.

So because the path difference is the same for both wavelength, since S1 to A and S2 to A does not change no matter what the wavelength is (remember A is fixed on the wall). This means we can equate n1L1 = n2L2. In this case both n1 and n2 are WHOLE numbers.

So looking at this equation, as L1 decreases to L2, n1 must INCREASE to n2.

So that means for L2 (the decreased wavelength) the whole number in front of the L2 is larger than the whole number in front of L1. So that means for the same distance from the middle line to A, there are more bright/dark bands on the wall for L2 than there are for L1, hence the the pattern of bright/dark bands get narrower (more tightly packed) for L2 (lower wavelength)

Now case 2: Assume that wavelength L1 and L2 produce a point at A which is a dark band. Due to destructive interference.

This is basically the same as case 1 but just let n1' = n1+0.5 and n2' = n2+0.5 Both n1 and n2 are whole numbers, hence n1' and n2' are positive integers.

so we get n1'L1 = n2'L2 hence using the same argument in case 1, the result is the same.

Now case 3: Assume that wavelength L1 produces a light band but L2 produces a dark band. (Or vice versa ofcourse)

So let the positive integer (ignoring 0 because that's middle line) in front of L2 be n2' = n2 + 0.5. Obviously n1 is just a whole number. n2 is a whole number which makes n2' a positive integer.

Using same argument as case 1, since Path difference is the same for both wavelength we can equate n1L1 = n2'L2.

From the same reasoning as L1 decrease to L2, n1 must INCREASE to n2' hence there is a larger positive integer in front of L2, hence it is narrower(more tightly packed)

In conclusion,  by showing that the top half of the central maximum is narrow then by symmetry the same can be shown for the bottom half, so by decreasing the wavelength we narrow the pattern produced on the wall.

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That's the end of the proof, is this correct? Or was what I did totally wrong?

Thank you!!!

[kamil9876]

Yes, or just learn the formula . Jk, nice stuff. Nice to see these sorts of proofs and enthusiasm for them in vce physics

[TrueTears]

Yes, or just learn the formula . Jk, nice stuff. Nice to see these sorts of proofs and enthusiasm for them in vce physics

ahhh ok, thanks for that kamil!!

Now I wonder what would happen if the points at A was a partial (as in not a bright or dark band...) hehehe, or is that not in the VCE course?