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King
 4 years 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
King
 4 years 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

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King
 4 years ago
Best ResponseYou've already chosen the best response.0@amistre64 @Callisto please help...

King
 4 years ago
Best ResponseYou've already chosen the best response.0@Mimi_x3 @pratu043 @Diyadiya help

King
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus @experimentX help..

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2can you draw the picture??

King
 4 years ago
Best ResponseYou've already chosen the best response.0they havent given one but wait i'll try...

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1So, okay, I got wire A 'square' and wire B 'circle' right?dw:1342104894718:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2are those independent??

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

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1yes, radius 'r' for loop 'B' = L/2pi

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

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

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1Oh yes I missed the '2pi', am sorry please account for that.

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

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

King
 4 years ago
Best ResponseYou've already chosen the best response.0so we get pi^2/16 but answer is pi ^2 /8sqrt2

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.1Now 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?...

King
 4 years ago
Best ResponseYou've already chosen the best response.0even i got this answer i was wondering hw this is wrong...... :(

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

King
 4 years ago
Best ResponseYou've already chosen the best response.0yeah its not a book its a worksheet..okay so thnx didi!!

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

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2what's the answer anyway??

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2magnetic 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
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