EAMCET Engineering 2006 Solved Paper

Considering the larger loop to be made up of four rods each of length L, the field at the centre, i.e., at a distance (‌

L

2

) from each rod, will beB‌=4×‌

µ0

4π

‌

l

d

[sin‌α+sin‌β]
‌ i.e., ‌‌‌B‌=4×‌

µ0

4π

‌

I

(‌ L 2 )

×2sin‌45
‌ i.e., ‌‌‌B1‌=‌

µ0

4π

‌

8√2

L

I
So, the flux with smaller loop
φ2=B1S2=‌

µ0

4π

‌

8√2

L

l2I
and hence, M=‌

φ2

I

=2√2‌

µ0

π

‌

l2

L

or ‌‌M∝‌

l2

L