AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

??

The jan2010 has everything on power losses in a transformer. Another paper has stuff on power transmission via national grid. Then you just need to know the basics like how a transformer works and that's all I think.

For convenience let the charges be x and y and let $k=\frac{1}{4\pi \varepsilon_0}$ (because we suspect we won't need the exact value since this is essentially a "ratios" exercise. By Coulomb's law we have $F=\frac{kxy}{d^2}$. We are also given that $0.5F=\frac{kxy}{20^2}$ (where all distances are in mm). Therefore substituting the first equation into the second one we get $0.5(\frac{kxy}{d^2})=\frac{kxy}{20^2}$, hence $\frac{0.5}{d^2}=\frac{1}{20^2}$. Now take square roots and stuff and solve for d.

EDIT: Whoops silly mistake, now edited.

By definition $F=Eq=1.5\times 10^5\times 1.6\times 10^{-19}=2.4\times 10^{-14}$, hence $F=ma=2.4\times 10^{-14}$. Since this is an electron $m=9.11\times 10^{-31}$. Plug in and solve to get a.

EDIT: Sniped

EQ=ma

Now solve as you know the charge and mass of an electron!


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19) Answer: D.
By Fleming's left hand rule the force acts at right angles to both the conventional current (opposite the velocity of the electron) and the field. Thus at any rate the force is at right angles to the field.

21) Answer: A.
$\Phi=BAN\cos \theta$.
At $\theta=50^o$ we have $\Phi=(2.8\times 10^{-2})(4.2\times 10^{-3})(50)(\cos 50^o)=3.779... Wb\ turns$.
At $\theta=0^o$ we clearly just have $\Phi=BAN$, which is simply the previous answer divided by $\cos 50^o$ (since previous answer = $BAN\cos 50^o$). Therefore $\frac{3.779...}{\cos 50^o}=5.88\times 10^{-3}Wb\ turns$.
It follows that $\Delta\Phi=5.88\times 10^{-3}-3.779...\times 10^{-3}=2.1\times 10^{-3}Wb\ turns\ (2\ s.f)$, as claimed.

24) Answer: Not entirely sure about this one but I think it's A.
Step up transformer so $N_p<N_s$. For any transformer at all it's obvious that $P_{out}\le P_{in}$ (with equality if and only if it is 100% efficient). Therefore $I_sV_s\le I_pV_p$ or $\frac{I_s}{I_p}\le \frac{V_p}{V_s}=\frac{N_p}{N_s}<1$ where the final inequality follows from the fact that $N_p<N_s$ (see above). We also have $\Phi\propto N$ since B,A and theta are all constant in this situation (and $\Phi=BAN\cos \theta$). Therefore $\frac{\Phi_s}{\Phi_p}=\frac{N_s}{N_p}>1$.

I got 23 in that one! One of my better ones


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Two pendulums, P and Q, are set up alongside each other. The period of P is 1.90s andthe period of Q is 1.95s. How many oscillations are made by pendulum Q between two consecutive instantswhen P and Q move in phase with each other?

any quick method for working these out?

19) Answer: D.
By Fleming's left hand rule the force acts at right angles to both the conventional current (opposite the velocity of the electron) and the field. Thus at any rate the force is at right angles to the field.

21) Answer: A.
$\Phi=BAN\cos \theta$.
At $\theta=50^o$ we have $\Phi=(2.8\times 10^{-2})(4.2\times 10^{-3})(50)(\cos 50^o)=3.779... Wb\ turns$.
At $\theta=0^o$ we clearly just have $\Phi=BAN$, which is simply the previous answer divided by $\cos 50^o$ (since previous answer = $BAN\cos 50^o$). Therefore $\frac{3.779...}{\cos 50^o}=5.88\times 10^{-3}Wb\ turns$.
It follows that $\Delta\Phi=5.88\times 10^{-3}-3.779...\times 10^{-3}=2.1\times 10^{-3}Wb\ turns\ (2\ s.f)$, as claimed.

24) Answer: Not entirely sure about this one but I think it's A.
Step up transformer so $N_p<N_s$. For any transformer at all it's obvious that $P_{out}\le P_{in}$ (with equality if and only if it is 100% efficient). Therefore $I_sV_s\le I_pV_p$ or $\frac{I_s}{I_p}\le \frac{V_p}{V_s}=\frac{N_p}{N_s}<1$ where the final inequality follows from the fact that $N_p<N_s$ (see above). We also have $\Phi\propto N$ since B,A and theta are all constant in this situation (and $\Phi=BAN\cos \theta$). Therefore $\frac{\Phi_s}{\Phi_p}=\frac{N_s}{N_p}>1$.

How is everyone spending tomorrow revising? It's my first full day before the exam focusing on PHYA4 -,-


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I'm lost...

In my chemistry exam :/ which is also in the afternoon, for bonus annoyingness for physics! And I have come down the the cold from hell so hopefully the revision I've done so far will be enough with a quick recap tomorrow evening

I think we need to know about light, heavy, critical and overdamping. It's good to know about overdamping even if it's possibly not on the spec (think about polar orbits not being on the spec but you still get questions about them here and there).

Light: The amplitude of oscillation reduces slowly over time.
Heavy: Amplitude of oscillation reduces more quickly over time than for a lightly damped system
Critical: The system doesn't oscillate but returns to its equilibrium position in the shortest possible time (in less than one cycle, due to no oscillation)
Overdamping: The system doesn't oscillate but also doesn't return to its equilibrium position. This is why you would perhaps not want to overdamp a car suspension system (car won't return to equilibrium, which is bad). You can think of this as a pendulum moving through a thick gooey substance from maximum displacement

Those are the general ideas anyway.