AQA Physics PHYA4 - 20th June 2016 [Exam Discussion Thread]

Yes - whenever you have to talk about the direction of a charge in the magnetic field you can say "Because of Flemings left hand rule it will in go in X direction (or the opposite if its negatively charged"

Right hand rule he meant, that's left :') don't think iv ever seen right come up in an exam tbh

Reply 701

boyyo

6

Original post by alevelstresss

if anyone wants, here are my PHYA4 notes :smile:

Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?

Reply 702

alevelstresss

2

Original post by boyyo

Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?

thats the emf of the cell

Reply 703

xMillnsy

1

Original post by alevelstresss

Find the speed at which the water moves

speed is equivalent to a length per second, so speed = volume / area

mass = density x volume

momentum = mass x velocity

Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?

Reply 704

Lychee628

19

Original post by SirRaza97

Like this. Sorry its in landscape.

Would it look this is if the right one was positive and the left one was negatively charged?

Reply 705

particlestudent

13

Original post by Ayaz789

So it'll look like this?

You haven't attached a picture :smile:

.

Reply 706

PiTheta97

2

Can someone show me their working for why the answer is B? Thanks
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1466345702365.jpg45.0KB

Reply 707

callumjmac

Original post by particlestudent

They are both positive.

Check page 22, I kind of explained it there :smile:

.

https://kaiserscience.files.wordpress.com/2015/04/electric-field-line-patterns.gif
Would it not be the top right image?

Reply 708

alevelstresss

2

Original post by xMillnsy

Still getting it wrong. Is volume cross-sectional area x the rate per second? so 7.2x10^-4 x 2x10^-4 = 1.44x10^-7 ?

length = volume / area
length = (2.0x10^-4)/(7.2x10^-4)
length = (5/18) m

length per second is equivalent to a velocity

v = (5/18) ms^-1

mass = volume x density
mass = 1000(2.0x10^-4)
mass = 0.2 kg

m = (1/5) kg

therefore using p = mv

p = (1/5) x (5/18)
p = 1/18
p = 0.056 kg ms^-1

Reply 709

callumjmac

Original post by boyyo

Hey, really useful notes thanks for these. You see in the capcitor section, it says E=Vc + Vr. What does the E(more of an epsilon i think) mean?

EMF of the power source in the circuit.

Reply 710

particlestudent

13

Original post by PiTheta97

Can someone show me their working for why the answer is B? Thanks
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Final flux linkage-initial

BAN-BANcos(50)

Reply 711

PiTheta97

2

Original post by particlestudent

Final flux linkage-initial

BAN-BANcos(50)

Made the mistake of just calculating BANcos50. Cheers

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Reply 712

callumjmac

Original post by PiTheta97

Can someone show me their working for why the answer is B? Thanks
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Flux linkage is greatest when the coil is perpendicular to the field lines.

When theta=50, Flux Linkage= BAN cos(theta) = 3.8x10^-3
When theta =0, Flux linkage = 5.9x10^-3

5.9x10^-3 - 3.8x10^-3 = 2.1 x10^-3 Wb Turns

Reply 713

xMillnsy

1

Original post by alevelstresss

length = volume / area
length = (2.0x10^-4)/(7.2x10^-4)
length = (5/18) m

length per second is equivalent to a velocity

v = (5/18) ms^-1

mass = volume x density
mass = 1000(2.0x10^-4)
mass = 0.2 kg

m = (1/5) kg

therefore using p = mv

p = (1/5) x (5/18)
p = 1/18
p = 0.056 kg ms^-1

ahh I get it now, thanks for your help

Reply 714

SirRaza97

7

Original post by Ayaz789

Haha its fine, draw it again and show me?

Like this? The repulsion area is closer to the "smaller" charge.

Reply 715

Lychee628

19

Original post by particlestudent

You haven't attached a picture :smile:

.

Well is the picture you attached on page 22?

Reply 716

abro1089

2

hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?

Reply 717

philo-jitsu

2

Original post by abro1089

hey physics people if I have two planets and want to work out the potential at a point between the two planets what do i do? Calculate the potentials due to each planet separately and add right? Also in the formula book gravitational potential is negative, do you sum the magnitude, or sum the negatives to get an even more negative value. And if the point is in the centre between the two planets?