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

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/AQA-PHYA4-2-W-QP-JUN11.PDF

(edited 8 years ago)

Reply 2022

OP

Original post by saad97

Can anyone help with this question please? Thanks.

[br]\text{Mass of container}=0.65kg[br]

The balance will read:

[br]0.65kg + \text{Mass of sand after 10s}[br]+ \text{Mass equivalence of force due to sand falling}[br]

[br]\text{Mass of sand after 10s} = [br]

[br]0.3kgs^{-1} \times 10s[br]

[br]= 3.00kg[br]

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

[br]= \dfrac{\text{Momentum per second}}{9.81}[br]

[br]= \dfrac{1.68}{9.81}[br]

[br]= 0.17kg[br]

Therefore, the balance reads:

[br]0.65kg + 3.00kg + 0.17kg[br]

[br]= 3.82kg[br]

(edited 8 years ago)

Reply 2023

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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Reply 2024

OP

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/AQA-PHYA4-2-W-QP-JUN11.PDF

ImageUploadedByStudent Room1432748946.964929.jpg

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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Reply 2025

OP

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?

Posted from TSR Mobile

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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Reply 2026

chughes17

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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Reply 2027

Adangu

Original post by CD223

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.

Posted from TSR Mobile

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.

Reply 2028

OP

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

Posted from TSR Mobile

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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Reply 2029

OP

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 :smile:

Reply 2030

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

ImageUploadedByStudent Room1432751320.649840.jpg

Reply 2031

OP

Original post by Jimmy20002012

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

ImageUploadedByStudent Room1432751320.649840.jpg

[br]E = \dfrac{\Delta V}{\Delta d}[br]

[br]E = \dfrac{30 - 20}{0.5}[br]

[br]E = 20 Vm^{-1}[br]

Reply 2032

Original post by CD223

[br]E = \dfrac{\Delta V}{\Delta d}[br]

[br]E = \dfrac{30 - 20}{0.5}[br]

[br]E = 20 Vm^{-1}[br]

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

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Reply 2033

Adangu

Original post by Jimmy20002012

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

Posted from TSR Mobile

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.

Reply 2034

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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Reply 2035

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, ImageUploadedByStudent Room1432754720.594658.jpg

ImageUploadedByStudent Room1432754720.594658.jpg

ImageUploadedByStudent Room1432754729.885328.jpg

Reply 2036

OP

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, ImageUploadedByStudent Room1432754720.594658.jpg

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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Reply 2037

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.

Posted from TSR Mobile

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

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Reply 2038

OP

Original post by Jimmy20002012

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

Posted from TSR Mobile

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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Reply 2039

gcsestuff