A square loop of side `a=6 cm` carries a current `I=1A`. Calculate magnetic induction `B(muT)` at point P, lying on the axis of loop and at a distance `x=sqrt7 cm` from the centre of loop.

Answer

9

Explanation

Therefore, perpendicular distance of P from each side of the loop is, `r=sqrt(x^(2)+(a/2)^(2))=4cm`
Now, consider only one side AB of the loop as shown in the figure (b)
`tan alpha=tan beta=((a//2))/r=3/4`
`alpha=beta=tan^(-1)(3/4)=37^(@)`
Magnitude of magnetic induction at P, due to current in this side AB, is


`B_(0)=(mu_(0)I)/(4pir)(sin alpha+sin beta)=3xx10^(-6)T`
Now, consider magnetic induction, produced by currents in two opposite side AB and CD as shown in the figure

Components of these magnetic inductions, parallel to plane of loop neutralise each other. Hence, resultant of these two magnetic inductions is `2B_(0)cos theta` (along the axis).
Similarly, resultant of magnetic inductions produced by currents in remaining two opposite sides BC and AD will also be equal to `2B_(0)cos theta` (along the axis in same direction).
Hence, resultant magnetic induction,
`B=4B_(0)cos theta, B=4xx(3xx10^(-6))xx((a//2))/r=9xx10^(-6)T=9muT`