## 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

1. anonymous

2. anonymous

3. anonymous

@UnkleRhaukus @experimentX help..

4. anonymous

@Mani_Jha

5. anonymous

@amistre64 @Callisto

6. anonymous

@.Sam. @apoorvk

7. anonymous

???????

8. experimentX

can you draw the picture??

9. anonymous

they havent given one but wait i'll try...

10. anonymous

So, okay, I got wire A 'square' and wire B 'circle' right?|dw:1342104894718:dw|

11. anonymous

yeah...

12. experimentX

are those independent??

13. anonymous

and R=L/2pi ?rite?

14. anonymous

Now, for wire A, magnetic field at 'O' will be the combined effect due to the four current carrying sides of the square loop.

15. anonymous

yeah....

16. anonymous

yes, radius 'r' for loop 'B' = L/2pi

17. anonymous

Now, magnetic field for A at O = 4 x $$\mu_oI/(L/8)$$

18. anonymous

isnt it mu i /2pir

19. anonymous

That's the square loop, (according to the formula -> B = $$\mu_oI/2\pi l$$

20. anonymous

Oh yes I missed the '2pi', am sorry please account for that.

21. anonymous

Now for the circular loop, $B= \mu_oI/2r\ = \mu_oI/2(L/2\pi) = \pi\mu_oI/L$

22. anonymous

And for the square loop, we found it out to be --> 4 x $$\mu_oI/2\pi(L/8)$$ = 16$$\mu_oI/\pi L$$

23. anonymous

so we get pi^2/16 but answer is pi ^2 /8sqrt2

24. anonymous

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?...

25. anonymous

even i got this answer i was wondering hw this is wrong...... :(

26. anonymous

:__(

27. anonymous

Lol there may just be a printing error in the book. Or. Did you copy out the question correctly?

28. anonymous

yeah its not a book its a worksheet..okay so thnx didi!!

29. anonymous

DIDI??? o.O *facepalm*

30. anonymous

:D

31. anonymous

Arey Sirji, I am a guy! (ab Bhaiya bolne ki zarurat nahi hai lekin :P )

32. anonymous

And whatever worksheet or book, please do check out what's wrong and let us know. :]

33. experimentX

34. anonymous

@experimentX I seem to get (pi^2)/16, but his book says (pi^2)/(8sqrt2).

35. experimentX

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