Find the experssion for the magnetic field at the centre O of a coil bent in the form of a square of side `2a`, carrying current I, figure.

Explanation

Here, `OE=AE=a`, `/_AOE=/_BOE=45^@`
Magnetic field at O due to current in arm AB of wire
`B_1=(mu_0)/(4pi)I/a[sin 45^@+sin45^@]=(sqrt2mu_0I)/(4pia)`
It is acting vertically downwards.
The magnetic field at O due to currents in arms BC, CD and DA will also be of the same magnitude and direction as `vecB_1`. Therefore, resultant magnetic field at O is
`B=4B_1=(4xxsqrt2mu_0I)/(4pia)=(sqrtmu_0I)/(pia)`
It is acting vertically downwards.