# A2 Physics questions - need help

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#1
Hi, could someone please explain why in http://qualifications.pearson.com/co...e_20110621.pdf

2 is A

and also why in http://qualifications.pearson.com/co...e_20110127.pdf

3 is B. Thanks very much in advance.
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5 years ago
#2
(Original post by Omghacklol)
Hi, could someone please explain why in http://qualifications.pearson.com/co...e_20110621.pdf
Q2. This is an application of Flemmings left hand rule (motors).

Instructions:

a) Pick a position at any place on the circular part of the conductor.

b) Note the direction of (conventional) current flow (+ve to -ve) at that place.

c) contort your hand so that the magnetic field lines and the current flowing align (tangent to the circle) with the appropriate fingers.

d) notice the force direction indicated by the remaining finger (thumb) will be towards the centre of the circle.

e) Now choose a different position 180 degrees displaced from the first and notice also that the force direction is towards the centre of the circle.

The conclusion is that the force will always be directed towards the centre of the circle over the whole circumference which therefore provides a compression force attempting to reduce the diameter of the circle.

You can make reaching the final conclusion a little faster by first eliminating one or two of the other options by the same left hand rule.
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5 years ago
#3
(Original post by Omghacklol)
Hi, could someone please explain why in http://qualifications.pearson.com/co...e_20110621.pdf

2 is A

and also why in http://qualifications.pearson.com/co...e_20110127.pdf

3 is B. Thanks very much in advance.
For number 2, the direction of the magnetic field perpendicular to the current flow at any point in the circular loop and if you use flemings left hand rule it will show the force produced acts in wards and hence if the switch is closed and there's a constant current flow which is perpendicular to the magnetic field there will be an inwards force all way round on the loop... hope this explanation makes sense For number 3, because E=V/d and E (field strength) is constant in uniform fields as the field lines are all in the same direction V/d should give the same value. So the graph of V against d, its gradient is E so it should be a constant gradient which in this case is B...Hope this helps 0
5 years ago
#4
(Original post by Omghacklol)
......and also why in http://qualifications.pearson.com/co...e_20110127.pdf

3 is B. Thanks very much in advance.
The question tells you the electric field is uniform. i.e. it is the same at all places between the plates.

We know that voltage is defined as Joules per coulomb of charge.

i.e. a charged object (particle) placed within the field will experience a force dependent on the sign of the charge. But that force will be constant anywhere in the field and will be directed towards one plate and away from the other.

Therefore, the work potential of that charge will be greatest at one plate and diminish to zero at the other plate. And since the field is uniform, the force acting on the particle is the same at all places so the work potential must be a linear (straight line) function.

B is therefore the only answer that fits the above observation.
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#5
(Original post by MSB47) 2 was very clear, thank you.

For 3 my confusion was that I thought E=V/D so V=DE i.e. V is proportional to D so I didn't see why it was a straight line that didn't pass the origin, which your answers kind of handwavingly explained.. I've just worked out rigourously why:

V = W/Q

but W = integral of F dx, so V = W/Q = the integral of F/Q dx = integral of E dx since E=F/Q by definition. We are given that E is constant so we have integral of E dx = Ex + c, where x is the distance from the plate. Substituting the initial condition that when x = 0, V is the initial voltage say V0, we have:

V=Ex+V0 which explains the nonzero y intercept V0. Do you think this is valid?

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5 years ago
#6
(Original post by Omghacklol)
2 was very clear, thank you.

For 3 my confusion was that I thought E=V/D so V=DE i.e. V is proportional to D so I didn't see why it was a straight line that didn't pass the origin, which your answers kind of handwavingly explained.. I've just worked out rigourously why:

V = W/Q

but W = integral of F dx, so V = W/Q = the integral of F/Q dx = integral of E dx since E=F/Q by definition. We are given that E is constant so we have integral of E dx = Ex + c, where x is the distance from the plate. Substituting the initial condition that when x = 0, V is the initial voltage say V0, we have:

V=Ex+V0 which explains the nonzero y intercept V0. Do you think this is valid?

Yes i guess that could although i'm not too familiar with calculus in physics
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5 years ago
#7
(Original post by Omghacklol)
2 was very clear, thank you.For 3 my confusion was that I thought E=V/D so V=DE i.e. V is proportional to D so I didn't see why it was a straight line that didn't pass the origin, which your answers kind of handwavingly explained.. I've just worked out rigourously why:V = W/Qbut W = integral of F dx, so V = W/Q = the integral of F/Q dx = integral of E dx since E=F/Q by definition. We are given that E is constant so we have integral of E dx = Ex + c, where x is the distance from the plate. Substituting the initial condition that when x = 0, V is the initial voltage say V0, we have:V=Ex+V0 which explains the nonzero y intercept V0. Do you think this is valid?Thanks very much for your answers though.
From your equation, as E > 0, as the distance from the plate increases, so does V. Take a look at the graph again.
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#8
(Original post by Phichi)
From your equation, as E > 0, as the distance from the plate increases, so does V. Take a look at the graph again.
If that's the case, how would you explain the shape of the graph?
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