## King Group Title 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 2 years ago 2 years ago

1. King

2. King

3. King

@UnkleRhaukus @experimentX help..

4. King

@Mani_Jha

5. King

@amistre64 @Callisto

6. King

@.Sam. @apoorvk

7. King

???????

8. experimentX

can you draw the picture??

9. King

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

10. apoorvk

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

11. King

yeah...

12. experimentX

are those independent??

13. King

and R=L/2pi ?rite?

14. apoorvk

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. King

yeah....

16. apoorvk

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

17. apoorvk

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

18. King

isnt it mu i /2pir

19. apoorvk

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

20. apoorvk

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

21. apoorvk

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

22. apoorvk

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. King

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

24. apoorvk

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. King

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

26. King

:__(

27. apoorvk

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

28. King

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

29. apoorvk

DIDI??? o.O *facepalm*

30. King

:D

31. apoorvk

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

32. apoorvk

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

33. 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