The physics a2 megathread

Original post by I'msoPi

Can someone help me out I keep getting like 6.3K which isn't even an option. How would I do this, seems like an awful amount of work for just one mark

Ill just draw the answer in paint. 2 secs

Reply 85

https://www.dropbox.com/s/7943hry43snz7y4/sss.JPG?dl=0

Original post by I'msoPi

Can someone help me out I keep getting like 6.3K which isn't even an option. How would I do this, seems like an awful amount of work for just one mark

Reply 86

https://www.dropbox.com/s/7943hry43snz7y4/sss.JPG?dl=0

Original post by Anika Sabah

Can someone please explain to me how to do this question.it would be a great help.thanks
How fast would a proton need to travel in order for the electromagnetic force on it to be sufficient to make it orbit the earth at the surface (radius of earth= 6.4*10^-19) comment on your answer.
(i was going to use F=B.Q.V.Sinx coz we have the charge of proton=1.6*10^-19 .but then we dont have B,magnetic field ). Please can someone help.tnx

What exam board is that from?

Reply 87

I'msoPi

13

Original post by clausencallum

It's AQA physics paper 2 specimen set 2

The answer says B - 8.7K

At least you got one of the options so you're basically on the right track
How come you did not take the latent heat of ice into consideration?

Reply 88

Original post by I'msoPi

It's AQA physics paper 2 specimen set 2

The answer says B - 8.7K

At least you got one of the options so you're basically on the right track
How come you did not take the latent heat of ice into consideration?

https://www.youtube.com/watch?v=xN-Q_mpk7mc

1:02

Reply 89

I'msoPi

13

Original post by clausencallum

https://www.youtube.com/watch?v=xN-Q_mpk7mc

1:02

Yeah I saw the video before. He does it in a fairly long way but I have it now :smile:

Reply 90

igneus1

3

You can't use mc(deltaT) for melting- you have to use latent heat stuff. mc(deltaT) only works when all the energy goes into heating the object (so no bonds broken). Presumably that's why 13.5 was a possible answer- to catch ppl out.

Reply 91

Original post by Raheel313

Each loop of wire has diameter 9.4 cm and the separation of the loops is 0.75 cm.

The coil is connected into a circuit such that the lower end of the coil is free to move.
(i) (i) Explain why, when a current is switched on in the coil, the separation of the loops of the coil decreases. [4]
(ii) Each loop of the coil may be considered as being a long straight wire.
In SI units, the magnetic flux density B at a distance x from a long straight wire carrying a current I is given by the expression
B = (2.0 × 10–7) I / x

When the current in the coil is switched on, a mass of 0.26 g is hung from the free end of the coil in order to return the loops of the coil to their original separation. Calculate the current in the coil. [4]
Help is needed in the second part. How do we know the current in the coil makes an angle of 90° with the magnetic field?

Current will flow in the first coil. It creates a magnetic field and there will be concentric circles of line of force in the vertical plane. The current in the 2nd coil is normal to the magnetic field created by the 1st coil and experiences a force. By N3L, the forces are equal and opposite and directed towards each other so separation decreases.

ii) mg = magnetic force

(edited 5 years ago)

Reply 92

cj104

11

Hi, I wanted to ask about the signal to noise ratio. If we have an optic fibre and want to obtain the signal to noise ratio at the receiver, do we use the input single power at the start of the fibre or the output signal power at the end?
I’ve come across two exercises and one of them uses the output and the other one the input

Reply 93

Original post by Raheel313

The second part in the mark scheme is solved by using the formula, F=BILsinθ and its further simolified to F=BIL, meaning the angle between the current in the coil and magnetic field is equal to 90°. I mean to ask that how do we know its 90? How is it visible from the diagram?

The first part is done by explaining the separation of the coils.

Current will flow in the first coil. It creates a magnetic field around it and there will be concentric circles of line of force in the vertical plane. The current in the 2nd coil is normal to the magnetic field created by the 1st coil and experiences a force. By N3L, the forces on each loop are equal and opposite and directed towards each other so separation decreases.

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html
This shows it

The loop is horizontal and the magnetic field forms concentric circles in the vertical plane.

This helps you in the second part.

mg=BIL

Reply 94

Original post by Raheel313

Plus, in the equation (B=2.00×10^-7×I÷x)there is this "variable" x, which is the distance from the long straight wire at which the magnetic Field Induction is equal to (B=2.00×10^-7×I/x). Why is x equal to the 0.75 cm which is the separation of both the loops?

Did you click this
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magcur.html
It shows field due to straight current carrying conductor

You assume the coil to be like long straight wire so you can use the above

x is the distance between the conductors

Current moves in the same direction in both loops

Reply 95

Original post by Raheel313

What conductors? Didnt we assume the whole coil to be a single, straight wire?

distance between first loop and second loop

yes

[quote me]

Reply 96

Original post by Raheel313

So, we are similarly treating both the loops as two different conductors?

Yes

Reply 97

Original post by Raheel313

So, in part (i) the current travels in the first loop which causes its magnetic field to act on the second loop, hence creating a force when current travels in the second loop. My question is, when the current enters the second loop wouldnt it no longer be in the first loop hence removing its magnetic field on the second loop? Or the current can be in both loops together? That way, when the current is in the second loop its magnetic field would act on the first loop creating a force the same way

According to your logic, if you have 2 resistors in series, when a current flows through the second, there is no current through the first. This can't be

There is a current flowing through both loops.

(edited 5 years ago)

Reply 98

Original post by Raheel313

One last question. When we have the case of 2 parallel current carrying conductors, we can easily deduce the magnetic field B of acting on one cuts normal to its current, hence determining the angle for use in formula. How can we identify this in cases like the one of the 2-looped coil given in this question?

It is the same principle

You consider 2 straight long wires

you use the right hand rule (Maxwell screw rule) to find direction of field.

You find that it is perpendicular to the current.

so

B-field due to 1 loop is normal to current in other loop

Reply 99