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.

Answer

`(9)`

Explanation

(9) Axis of the loop means a line passing through centre of loop and
normal to its plane. Since distance of the point P is x from centre
of loop and side of square loop is a as shown in Fig


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
Fig.
`tan alpha=tan beta=(a//2)/r=3/4`
`alpha=beta=tan^-1(3/4)=37^@`
Magnitude of magnetic field induction at P, due to current in this side
AB, is
`B_0=(mu_0I)/(4pir)(sin alpha+sin beta)=3xx10^-6T`
Now, consider magnetic induction, produced by currents in two
opposite sides AB and CD as shown in Fig.

Component of these mangetic inductions parallel to plane of
loop neutralise each other. Hence, resultant of these two magnetic
induction is `2B_0cos 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_0cos theta` (along the axis in same direciton).
Hence, resultant magnetic induction,
`B=4B_0 cos theta`
`B=4xx(3xx10^-6)(a//2)/r=9xx10^-6T=9muT`