Motors And Genertators Is Harddd!!! (1 Viewer)
- Thread starter aero135
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Pwnage101
Moderator
i am
in fact i have the book, so just put the page reference up
in fact i have the book, so just put the page reference up
Pwnage101
Moderator
Okay i'll do 'em now:
1.4.1 (b)
okay - from 1.4.1 (a), you get 2.5 x 10^-6 N attraction
now the forumula they give you in the formula sheet for HSC is F/l = etc...
now they give it in that form for a reason, because that is used to calculate 'force per unit length', ie the force for every 1 m
so you just divide your answer of (a) by the length to get teh force per unit length, which is (2.5 x 10^-6)/(0.3) = 8.3 x 10^-6 N attraction
so the answer in the back is Correct - note this is a very simple question and you are required to understand how to do this
1.4.2
okay all the asnwers are correct (from what i've calculated) other than that for wire R, which i got 6.67 x 10 ^-6 (when the answer is very close, 0.69 x 10^-5
so basically for wire P, wwire Q and wire R ar eboth exerting a force on it
wire Q is pushing up, while wire R is pulling it down the page - you need to calculate teh force each is exerting,a nd subtract them to find teh net force acting, post up if you have trouble doing this (it shouold be very straight forward - note qire R and wire P are 0.3 m apart (0.1 + 0.2))
1.4.3
(a) - the diagram is correct, so basically u apply the right hand rule and note that the top wire will have circles on top o it, crosses beneath, while the wire beneath it (which has I going the otha way) has crosses on top and circles beneath
(b) the explanation they give is alright,m but here's how i would explain it:
now call the top wire wire 1 and the bottom wire wire 2. Now the magnetic field due to wire 1, according to teh right hand rule, goes into the page benath it and out of the page on top of it
since wire 2 is beneath it, the magnetic field from wire 1 is going into the page there. Applying the Right Hand Push Rule for teh force acting on wire 2 due to the current in wire 2 and teh magnetic field due to wire 1 - we point our thumb to the left, our finger into the page, and so the resulting force is downwards. Repeat the process for wire 1 using the magnetic field due to wire 2. That force will push up. So, as explained, each wire is forced away from the other,a nd thus there is repulsion
1.5.1
(a)
well, you take the current at any poiint as a tangent, and thus using right hand push rule, you get the force perpindicular outwards to this tangent, and so the answer is correct
(b)zero, as the answer correctly states, because the force pulling it down is perfectly compensated by teh force pulling it down, etc, since it is perfectly symmetrical (ie a circle)
jm01, yeh there are a couple of minor errors, but overall the book is great!
1.4.1 (b)
okay - from 1.4.1 (a), you get 2.5 x 10^-6 N attraction
now the forumula they give you in the formula sheet for HSC is F/l = etc...
now they give it in that form for a reason, because that is used to calculate 'force per unit length', ie the force for every 1 m
so you just divide your answer of (a) by the length to get teh force per unit length, which is (2.5 x 10^-6)/(0.3) = 8.3 x 10^-6 N attraction
so the answer in the back is Correct - note this is a very simple question and you are required to understand how to do this
1.4.2
okay all the asnwers are correct (from what i've calculated) other than that for wire R, which i got 6.67 x 10 ^-6 (when the answer is very close, 0.69 x 10^-5
so basically for wire P, wwire Q and wire R ar eboth exerting a force on it
wire Q is pushing up, while wire R is pulling it down the page - you need to calculate teh force each is exerting,a nd subtract them to find teh net force acting, post up if you have trouble doing this (it shouold be very straight forward - note qire R and wire P are 0.3 m apart (0.1 + 0.2))
1.4.3
(a) - the diagram is correct, so basically u apply the right hand rule and note that the top wire will have circles on top o it, crosses beneath, while the wire beneath it (which has I going the otha way) has crosses on top and circles beneath
(b) the explanation they give is alright,m but here's how i would explain it:
now call the top wire wire 1 and the bottom wire wire 2. Now the magnetic field due to wire 1, according to teh right hand rule, goes into the page benath it and out of the page on top of it
since wire 2 is beneath it, the magnetic field from wire 1 is going into the page there. Applying the Right Hand Push Rule for teh force acting on wire 2 due to the current in wire 2 and teh magnetic field due to wire 1 - we point our thumb to the left, our finger into the page, and so the resulting force is downwards. Repeat the process for wire 1 using the magnetic field due to wire 2. That force will push up. So, as explained, each wire is forced away from the other,a nd thus there is repulsion
1.5.1
(a)
well, you take the current at any poiint as a tangent, and thus using right hand push rule, you get the force perpindicular outwards to this tangent, and so the answer is correct
(b)zero, as the answer correctly states, because the force pulling it down is perfectly compensated by teh force pulling it down, etc, since it is perfectly symmetrical (ie a circle)
jm01, yeh there are a couple of minor errors, but overall the book is great!
Last edited:
Pwnage101
Moderator
i edited my post, but yeh as i said, a lot of the asnwers are solid