CGPET 2016 Solved Paper

Let O be the centre of square ABCD of side 2a carrying current i ampere. The magnetic field due to infinite length of wire AD is given by
B1=

μ0i

4πa

(sin‌α+sin‌β)
(Here,α=β=45°)

μ0i

4πa

(sin‌45°+sin‌45°)
=

μ0i

4πa

(

1

√2

+

1

√2

)
=

√2μ0i

4πa

So, by the symmetry the magnetic field at O due to each side will be

√2.μ0i

4πa

So, Net magnetic field at the centre O of current carrying square is,
B=4B1
=

4×√2μ0i

4π(a)

=√2

μ0i

πa