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sarbear.h

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can someone help with hsc 2010 q20
edit: btw i checked the physics high yt vid... they used the right hand curl rule which i totally agree with except with like answer A he said the left was the south???? is it not the north???
1742723042756.png
 

cheesynooby

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current flows for pos to neg so in (A) the coils of wire can be thought of as a solenoid that forms an electromagnet with N pole to the left. i.e. the B field across the armature is to the left.
following the + terminal leads to the right brush - so this is always +ve. because of the split ring commutator the right side of the armature is also always +ve. using the right hand rule, with I going into page, B to left, the F is up and the armature rotates anticlockwise.
B is the same as A except the brushes are switched so the armature rotates clockwise.
in C and D the right magnet and left electromagnet form north poles facing each other so there is no uniform B field and can be discarded.
edit: the left acts as the south pole for the armature because current flows up the side facing u and down the other side - so the N pole is on the left. thus the S pole is on the right of the magnet on the left which is what receives the B field experienced by the armature
 

alphxreturns

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first we want the magnetic field to run N to S, alternatively S to N (for uniform B field)

using the right hand screw rule for looped current, we see that the left magnet thing will have a magnetic field with North going left (thus south going right)

so, in the right magnet, we want a similar north-going-left magnetic field. use the right hand rule again to see that both C and D have a north-going-right field.

We are left with A and B, the difference between them being in which side to link the positive and negative terminals to.

as seen in the diagrams we want a certain rotation result: counter-clockwise

use the right hand palm rule, to, if the B-field was going right to left, see what direction of current would give us the right force to cause this rotation

ok nvm cheesynoby got this under control :music:
 

sarbear.h

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current flows for pos to neg so in (A) the coils of wire can be thought of as a solenoid that forms an electromagnet with N pole to the left. i.e. the B field across the armature is to the left.
following the + terminal leads to the right brush - so this is always +ve. because of the split ring commutator the right side of the armature is also always +ve. using the right hand rule, with I going into page, B to left, the F is up and the armature rotates anticlockwise.
B is the same as A except the brushes are switched so the armature rotates clockwise.
in C and D the right magnet and left electromagnet form north poles facing each other so there is no uniform B field and can be discarded.
edit: the left acts as the south pole for the armature because current flows up the side facing u and down the other side - so the N pole is on the left. thus the S pole is on the right of the magnet on the left which is what receives the B field experienced by the armature
first we want the magnetic field to run N to S, alternatively S to N (for uniform B field)

using the right hand screw rule for looped current, we see that the left magnet thing will have a magnetic field with North going left (thus south going right)

so, in the right magnet, we want a similar north-going-left magnetic field. use the right hand rule again to see that both C and D have a north-going-right field.

We are left with A and B, the difference between them being in which side to link the positive and negative terminals to.

as seen in the diagrams we want a certain rotation result: counter-clockwise

use the right hand palm rule, to, if the B-field was going right to left, see what direction of current would give us the right force to cause this rotation

ok nvm cheesynoby got this under control :music:
yeah bc thats what i thought like if ur using the right hand curl rule then for like A, the north will be left south be right
bruh phys high got me messed up
 

sarbear.h

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current flows for pos to neg so in (A) the coils of wire can be thought of as a solenoid that forms an electromagnet with N pole to the left. i.e. the B field across the armature is to the left.
following the + terminal leads to the right brush - so this is always +ve. because of the split ring commutator the right side of the armature is also always +ve. using the right hand rule, with I going into page, B to left, the F is up and the armature rotates anticlockwise.
B is the same as A except the brushes are switched so the armature rotates clockwise.
in C and D the right magnet and left electromagnet form north poles facing each other so there is no uniform B field and can be discarded.
edit: the left acts as the south pole for the armature because current flows up the side facing u and down the other side - so the N pole is on the left. thus the S pole is on the right of the magnet on the left which is what receives the B field experienced by the armature
u said that the split right commutator on the right side of the armature is always +ve? but answer B.. the pos side isnt connected to the right brush...? or is what u said only applicable for ans A

lowkey this still dont make sense.. i only get why A isn't the answer
 
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cheesynooby

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u said that the split right commutator on the right side of the armature is always +ve? but answer B.. the pos side isnt connected to the right brush...? or is what u said only applicable for ans A

lowkey this still dont make sense.. i only get why A isn't the answer
yeah in B, the pos and neg terminals connected to opposite brushes from A
i.e. pos on left brush, neg on right brush
B field is still to the left
and in the rotor coil the I goes clockwise so force is up on the left side and down on the right side -> clockwise
 

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