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

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    The mass of the container is 0.650 kg. Show that the reading of the balance, 10.0 s afterthe sand starts landing continuously in the container, will be 3.82 kg. You may assumethat the sand comes to rest without rebounding when it lands in the container.
    Jun 12 Q1biii.

    Can anyone help with this question please? Thanks.
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    Can someone please help me with question 4 (b)(i) on June 2011 PHYA4/2??

    I never understand the hand rule ones, for some reason they just don't make sense!

    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
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    (Original post by saad97)
    Can anyone help with this question please? Thanks.
    

\text{Mass of container}=0.65kg

    The balance will read:

    

0.65kg + \text{Mass of sand after 10s}

+ \text{Mass equivalence of force due to sand falling}

    

\text{Mass of sand after 10s} =

    

0.3kgs^{-1} \times 10s

    

= 3.00kg

    

\text{Mass equivalence of force due to sand falling} = \dfrac{\text{Force of sand}}{g}

    

= \dfrac{\text{Momentum per second}}{9.81}

    

= \dfrac{1.68}{9.81}

    

= 0.17kg

    Therefore, the balance reads:

    

0.65kg + 3.00kg + 0.17kg

    

= 3.82kg
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    June 11 MC Q23

    How do we know it INCREASES then stays the same and doesn't DECREASE then stay the same, i.e. Negative magnetic flux linkage?




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    (Original post by Adangu)
    Can someone please help me with question 4 (b)(i) on June 2011 PHYA4/2??

    I never understand the hand rule ones, for some reason they just don't make sense!

    http://filestore.aqa.org.uk/subjects...W-QP-JUN11.PDF
    Name:  ImageUploadedByStudent Room1432748946.964929.jpg
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    You want the magnetic force to be vertically upwards to keep the positive ion on its track undeflected and counter the force causing it to move downwards.

    The current is in the same direction as the path of the positive ions (as it is conventional current.

    Using Fleming's LHR, the field needs to be applied into the plane of the diagram for the magnetic force to act vertically upwards and the current to act in the same direction as the motion of the positive ions.


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    (Original post by chughes17)
    June 11 MC Q23

    How do we know it INCREASES then stays the same and doesn't DECREASE then stay the same, i.e. Negative magnetic flux linkage?




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    Think of the magnetic flux linkage as the "amount of flux" cut by N turns of a coil.

    If the magnet is pushed inside the coil, the flux due to the permanent magnet is increasingly cut by more and more turns of the coil, increasing the flux linkage.


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    (Original post by CD223)
    Think of the magnetic flux linkage as the "amount of flux" cut by N turns of a coil.

    If the magnet is pushed inside the coil, the flux due to the permanent magnet is increasingly cut by more and more turns of the coil, increasing the flux linkage.


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    So if North went in first then it'd still be an increase? Also, in which case how can there ever be negative flux linkage? Is that strictly only for rotating coils?

    Thanks


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    (Original post by CD223)
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    You want the magnetic force to be vertically upwards to keep the positive ion on its track undeflected and counter the force causing it to move downwards.

    The current is in the same direction as the path of the positive ions (as it is conventional current.

    Using Fleming's LHR, the field needs to be applied into the plane of the diagram for the magnetic force to act vertically upwards and the current to act in the same direction as the motion of the positive ions.


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    You my friend are an absolute lifesaver! I didn't understand the question properly but, the way you answered it made it so simple. I didn't realise I was trying to stop the deflection!

    Wish you were my teacher mate, really do. Teachers in my school aren't the best.
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    (Original post by chughes17)
    So if North went in first then it'd still be an increase? Also, in which case how can there ever be negative flux linkage? Is that strictly only for rotating coils?

    Thanks

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    I believe there would still be an increase, yes, because more and more turns of the coil are still cutting flux likes of the magnet.

    Negative flux linkage refers only to the orientation of a rotating coil.

    In theory, and strictly, the magnitude of the flux linkage is always positive for a rotating coil in a magnetic field of a permanent magnet (BAN \cos \theta), where theta is the angle between the normal of the coil face and the field lines.

    However, you sometimes see magnetic flux being written as negative to show the conventions of a sinusoidal variation in the coil's rotations.


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    (Original post by Adangu)
    You my friend are an absolute lifesaver! I didn't understand the question properly but, the way you answered it made it so simple. I didn't realise I was trying to stop the deflection!

    Wish you were my teacher mate, really do. Teachers in my school aren't the best.
    No worries! Glad you found it helpful


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    For some reason can't do this multichoice, can someone help?

    Name:  ImageUploadedByStudent Room1432751320.649840.jpg
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    (Original post by Jimmy20002012)
    For some reason can't do this multichoice, can someone help?

    Name:  ImageUploadedByStudent Room1432751320.649840.jpg
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E = \dfrac{\Delta V}{\Delta d}

    

E = \dfrac{30 - 20}{0.5}

    

E = 20 Vm^{-1}


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    Ohhhh. Though you could only use that equation it is in a uniform field, isn't this a radial field?


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    (Original post by Jimmy20002012)
    Ohhhh. Though you could only use that equation it is in a uniform field, isn't this a radial field?


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    This isn't a radial field as E would be inversely proportional to r^2.

    It is also a straight line and can be said to be uniform.
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    (Original post by Adangu)
    This isn't a radial field as E would be inversely proportional to r^2.

    It is also a straight line and can be said to be uniform.
    Ahh okay, thanks.


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    (Original post by Adangu)
    This isn't a radial field as E would be inversely proportional to r^2.

    It is also a straight line and can be said to be uniform.
    I am pretty sure this is a similar question. Why in the do they use the electric field strength equation for a radial field, part bii, Name:  ImageUploadedByStudent Room1432754720.594658.jpg
Views: 120
Size:  53.2 KBName:  ImageUploadedByStudent Room1432754729.885328.jpg
Views: 105
Size:  51.4 KB


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    (Original post by Jimmy20002012)
    I am pretty sure this is a similar question. Why in the do they use the electric field strength equation for a radial field, part bii, Name:  ImageUploadedByStudent Room1432754720.594658.jpg
Views: 120
Size:  53.2 KBName:  ImageUploadedByStudent Room1432754729.885328.jpg
Views: 105
Size:  51.4 KB


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    In the MC question, the giveaway is that the same increase in potential \Delta V is observed over equal distances \Delta d.



    In the written question there is no indication that the potential difference is equal over equal intervals.


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    (Original post by CD223)
    In the MC question, the giveaway is that the same increase in potential \Delta V is observed over equal distances \Delta d.



    In the written question there is no indication that the potential difference is equal over equal intervals.


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    So if the potential are equal, uniform field if not stated or not equal radial field right!?


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    (Original post by Jimmy20002012)
    So if the potential are equal, uniform field if not stated or not equal radial field right!?


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    If an equal potential differences are observed over equal distance intervals the field is uniform.

    I would make the assumption that when dealing with point charges, unless otherwise stated or indicated on a diagram, the field is radial.


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    Using flemings left hand rule, am i right in thinking the current finger is always in the direction of current, so for example if an electron is going right in a magnetic field, the conventional current and therefore my 2nd finger would be pointing left.
 
 
 
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