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King
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
@amistre64 @Callisto please help...
@Mimi_x3 @pratu043 @Diyadiya help
@UnkleRhaukus @experimentX help..
can you draw the picture??
they havent given one but wait i'll try...
So, okay, I got wire A 'square' and wire B 'circle' right?|dw:1342104894718:dw|
are those independent??
Now, for wire A, magnetic field at 'O' will be the combined effect due to the four current carrying sides of the square loop.
yes, radius 'r' for loop 'B' = L/2pi
Now, magnetic field for A at O = 4 x \(\mu_oI/(L/8) \)
That's the square loop, (according to the formula -> B = \(\mu_oI/2\pi l\)
Oh yes I missed the '2pi', am sorry please account for that.
Now for the circular loop, \[B= \mu_oI/2r\ = \mu_oI/2(L/2\pi) = \pi\mu_oI/L\]
And for the square loop, we found it out to be --> 4 x \(\mu_oI/2\pi(L/8)\) = 16\(\mu_oI/\pi L\)
so we get pi^2/16 but answer is pi ^2 /8sqrt2
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?...
even i got this answer i was wondering hw this is wrong...... :(
Lol there may just be a printing error in the book. Or. Did you copy out the question correctly?
yeah its not a book its a worksheet..okay so thnx didi!!
Arey Sirji, I am a guy! (ab Bhaiya bolne ki zarurat nahi hai lekin :P )
And whatever worksheet or book, please do check out what's wrong and let us know. :]
what's the answer anyway??
@experimentX I seem to get (pi^2)/16, but his book says (pi^2)/(8sqrt2).
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/Example-9-1-Magnetic-Field-due-to-a-Finite-Straight-Wire