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A square loop of side 6 cm carries a cur...

A square loop of side `6 cm` carries a current of `30 A`. Calculate the magnitude of magnetic field `B` at a point `P` lying on the axis of the loop and a distance`sqrt7` cm from centre of the loop.

Text Solution

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`AP=sqrt(x^2+(a^2)/4)=(sqrt(4x^2+a^2))/2`
`CP=sqrt(AC^2+AP^2)=sqrt(x^2+(a^2)/2)=(sqrt(4x^2+2a^2))/2`

`sin alpha= (AC)/(AP)=a/(sqrt(4x^2+2a^2))`
Now `B=(mu_0i)/(4pi(AP)) 2sin alpha=(mu_0ia)/(pisqrt(4x^2+a^2)sqrt(4x^2+2a^2))`
When the fields of all the four sides are considered, the horizontal
components add to zero. So, the total field is given by
`B_R = 4Bcos theta = (4mu_0ia^2)/(pi(4x^2+a^2) sqrt(4x^2+2a^2))`
For `x=0`, the expression reduces to `B_R = (4mu_0ia^2)/((pia^2)sqrt(2a^2))=(2sqrt2mu_0i)/(pia)`
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