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Kaido

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What answer did you get Fizzy?

JoshMosh2 argued that it was C (apparently this was the case in an independent exam) - he said that we don't use the sin30 and simply do F=3x10^-4x 3 x 0.3

So I'm extremely confused as I was under the assumption that F=3x10^-4x 3 x 0.3 x sin30

Initially, the exam had A as the answer, but after hearing what josh said, our teacher changed the answer to C. o.o
 

el_manu

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What answer did you get Fizzy?

JoshMosh2 argued that it was C (apparently this was the case in an independent exam) - he said that we don't use the sin30 and simply do F=3x10^-4x 3 x 0.3

So I'm extremely confused as I was under the assumption that F=3x10^-4x 3 x 0.3 x sin30

Initially, the exam had A as the answer, but after hearing what josh said, our teacher changed the answer to C. o.o

What did josh say?
 
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Kaido

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Well, Josh said that the length of wire is 30cm and thus we no longer need to multiply by sin30. He further backed his argument as he said that the answer (to an identical question) in an independent paper was C
 

el_manu

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That is weird indeed..


I don't know. In a case like this: 06q7nc7.gif



the wire is always perpendicular to the magnetic field, so theta = 90 degrees.....and sin90 = 1.
So in this case F = BIL is all you would need. However, in your case, I would think that you would need to times by sin(30)..

Sorry I'm not much help
 

123ryoma12

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Answer is A. Sin30 is required in this as the wire is not perpendicular to the magnetic field.
Just note that if the magnetic field is going into/out of the page then in most cases it will be perpendicular to the current unless the current is also moving in/out of the page.
 

Fizzy_Cyst

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What answer did you get Fizzy?

JoshMosh2 argued that it was C (apparently this was the case in an independent exam) - he said that we don't use the sin30 and simply do F=3x10^-4x 3 x 0.3

So I'm extremely confused as I was under the assumption that F=3x10^-4x 3 x 0.3 x sin30

Initially, the exam had A as the answer, but after hearing what josh said, our teacher changed the answer to C. o.o
Really?

No way. The answer should be A.

It states that the wire is of length 30cm, not that the width of the field is 30cm. If you were told that the width of the field was 30cm and the wire was within it, then fair enough, l should be 0.30m. But, seeing as the length of the wire is 30cm and it creates an angle from the perpendicular to the field, you need to solve for the perpendicular component as that is the only component which experiences a force.

This would be analogous (ok, not really analogous but EQUALLY INCORRECT) to saying that a projectile is launched at 30m/s at an angle of 30 degrees above horizontal and then using Uy = 30m/s.

Answer is A. JoshMosh (and your easily confused teacher) is (are) wrong. Soz m8.

Btw, independent marking schemes are not a valid way of determining if an answer is correct, there are always mistakes, every year. Now, tell old Joshy to find me a similar HSC question where they do not use the angle.
 
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Squar3root

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I got my answer to be A as well.

The wire is not perpendicular to the magnetic field so you do need to only include the component perpendicular to the magnetic field.

in the below question the wire is perpendicular to the magnetic field so you don't need to use the "sin" part. (however they trick you into thinking the length of the wire is 0.4m :haha:)

That is weird indeed..

I don't know. In a case like this: View attachment 32029

the wire is always perpendicular to the magnetic field, so theta = 90 degrees.....and sin90 = 1.
So in this case F = BIL is all you would need. However, in your case, I would think that you would need to times by sin(30)..

Sorry I'm not much help
I thought of this question too LOL
 

someth1ng

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Well, Josh said that the length of wire is 30cm and thus we no longer need to multiply by sin30. He further backed his argument as he said that the answer (to an identical question) in an independent paper was C
That's just wrong...LOL.
 

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