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2 Identical wires A and B have the same length L and carry the same current I.Wire A is bent to form a square of side a.B1 and B2 are the values of magnetic induction at the center of the circle and center of the square respectively.Find B1/B2
 one year ago
 one year ago
2 Identical wires A and B have the same length L and carry the same current I.Wire A is bent to form a square of side a.B1 and B2 are the values of magnetic induction at the center of the circle and center of the square respectively.Find B1/B2
 one year ago
 one year ago

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KingBest ResponseYou've already chosen the best response.0
@amistre64 @Callisto please help...
 one year ago

KingBest ResponseYou've already chosen the best response.0
@Mimi_x3 @pratu043 @Diyadiya help
 one year ago

KingBest ResponseYou've already chosen the best response.0
@UnkleRhaukus @experimentX help..
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
can you draw the picture??
 one year ago

KingBest ResponseYou've already chosen the best response.0
they havent given one but wait i'll try...
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
So, okay, I got wire A 'square' and wire B 'circle' right?dw:1342104894718:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
are those independent??
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Now, for wire A, magnetic field at 'O' will be the combined effect due to the four current carrying sides of the square loop.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
yes, radius 'r' for loop 'B' = L/2pi
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Now, magnetic field for A at O = 4 x \(\mu_oI/(L/8) \)
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
That's the square loop, (according to the formula > B = \(\mu_oI/2\pi l\)
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Oh yes I missed the '2pi', am sorry please account for that.
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Now for the circular loop, \[B= \mu_oI/2r\ = \mu_oI/2(L/2\pi) = \pi\mu_oI/L\]
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
And for the square loop, we found it out to be > 4 x \(\mu_oI/2\pi(L/8)\) = 16\(\mu_oI/\pi L\)
 one year ago

KingBest ResponseYou've already chosen the best response.0
so we get pi^2/16 but answer is pi ^2 /8sqrt2
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Now you can find out the ratio! \[\frac{B_A}{B_B} =\frac{16 μ_oI/πL}{πμ_oI/L} = \pi^2/16\] Now where are we going wrong hmmm?...
 one year ago

KingBest ResponseYou've already chosen the best response.0
even i got this answer i was wondering hw this is wrong...... :(
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Lol there may just be a printing error in the book. Or. Did you copy out the question correctly?
 one year ago

KingBest ResponseYou've already chosen the best response.0
yeah its not a book its a worksheet..okay so thnx didi!!
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
DIDI??? o.O *facepalm*
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
Arey Sirji, I am a guy! (ab Bhaiya bolne ki zarurat nahi hai lekin :P )
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
And whatever worksheet or book, please do check out what's wrong and let us know. :]
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
what's the answer anyway??
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
@experimentX I seem to get (pi^2)/16, but his book says (pi^2)/(8sqrt2).
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
magnetic field due to circular loop is \[ \vec B_1 = {\mu_0I \pi \over L}\] Magnetic field due to square loop is \[ \vec B_2 = {8\sqrt 2\mu_0 I \over L \pi}\] \[ B_1/B_2 = \pi^2/8 \sqrt 2\] http://www.scribd.com/doc/56083415/3/Example91MagneticFieldduetoaFiniteStraightWire
 one year ago
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