Find magnitude and direction of magnetic field at point `P` in the following cases.
(a)
(b)
`P` is the centre of square.
(c)
`P` is the centre of equilateral triangle.
(d)
`P` is the centre of regular hexagon.
(e)
`P` is the centre of rectangular loop.
(f)
(g) A long wire carrying a current `i` is bent to from a plane angle `theta`. Find magnetic field at a point on the bisector of this angle is situated at a distance `d` from vertex.

(h) A long, straight wire carriers a current `i`. Let `B_(1)` be the magnetic field at a point `P` at a distance `d` from the wire. consider a section of length `l` of this wire such that the point `P` lies on a perpendicular bisector of the sector. Let `B_(2)` be the magnetic field at this point due to this section only. find the value of `d//l` so that `B_(2)` differes from `B_(1)` by `1%`.

Explanation

(a)
`B_(P)=(mu_(0)i)/(4pid')[cos theta_(1)+cos theta_(2)]`
`=(mu_(0)i)/(4pi(3d))[cos theta+cos theta]`
`=(mu_(0)i)/(6pid) cos theta`
`cos theta=4/5`
`B_(P)=(mu_(0)i)/(6pid) .4/5=(2mu_(0)i)/(15pid) ox`.
(b)
Magnetic field at `P` due to one side of square, say `AB`
`B_(1)=(mu_(0)i)/(4pid')(cos theta+cos theta)`
`d'=d, theta=45^(@)`
`=(mu_(0)i)/(4pid).2.(1)/(sqrt(2))=(mu_(0)i)/(2sqrt(2)pid)`
Magnetic field due to square loop
`B_(p) = 4B_(1) = (sqrt2mu_(0)i)/(pi d)`
Since current anticlockwise, magnetic field at `P` will be outside the plane of paper.
`B_(P)=(sqrt(2)mu_(0)i)/(pid), o.`
(c)
`tan 30^(@) =(d')/(d//2) implies d'=d/(2sqrt(3))`
`theta_(1)=theta_(2)=30^(@)`
`B_(1)=(mu_(0)i)/(4pid')[cos30^(@)+cos30^(@)]`
`=(mu_(0)i)/(4pi.d/(2sqrt(3))).2 (sqrt(3))/2=(3mu_(0)i)/(2pid)`
Magnetic field at `P` due to triangular loop
`B_(P)=3B_(1)=(9mu_(0)i)/(2pid)`
Since current is clockwise, magnetic field at `P` inside the plane of paper.
`B_(p)=(9mu_(0)i)/(2pid)`
Since current is clockwise, magnetic field at `P` inside the plane of paper.
`B_(p)=(9mu_(0)i)/(2pid), ox`
(d)
`B_(1)=(mu_(0)i)/(4pid')(cos 60^(@)+cos 60^(@))`
`=(mu_(0)i)/(4pi.d/2tan 60^(@)).2 cos 60^(@)`
`=(mu_(0)i)/(2pidsqrt(3))`
`B_(P)=6B_(1)=(sqrt(3)mu_(0)i)/(pid), ox`
(e)
From `triangleADC,AC=10d`
`cos alpha =(8d)/(10d)=4/5, cos beta=(6d)/(10d)=3/5`
`d_(1)=3d, d_(2)=4d`
`B_(1)=(mu_(0)i)/(4pid_(1))(cos alpha+cos alpha)`
`=(mu_(0)i)/(4pi.3d)xx2xx4/5=(2mu_(0)i)/(15pid)`
`B_(2)=(mu_(0)i)/(4pid_(2))(cos beta+cos beta)`
`=(mu_(0)i)/(4pi.4d)xx2xx3/5=(3mu_(0)i)/(40pid)`
`B_(P)=2(B_(1)+B_(2))=2((2mu_(0)i)/(15pid)+(3mu_(0)i)/(40pid))`
`=(2mu_(0)i)/(pid)(2/15+3/40=(16+9)/120)`
`=(5mu_(0)i)/(12pid), ox`
(f)

Magnetic field at `P`
Due to `AB:`

`B_(1)=(mu_(0)i)/(4pid_(2))(cos 90^(@)+cos 45^(@))`
`=(mu_(0)i)/(4sqrt(2)pid), ox`
Due to `BC`

`B_(2)=(mu_(0)i)/(4pid_(2))(cos 45^(@)+cos 90^(@))`
`=(mu_(0)i)/(4sqrt(2)pid), ox`
Due to `CA`

`B_(3)=(mu_(0)i)/(4pid')(cos 45^(@)+cos 45^(@))`
`=(mu_(0)i)/(4pi.d/(sqrt(2)))xx2xx1/(sqrt(2))=(mu_(0)i)/(2pid), o.`
`B_(3)gt(B_(1)+B_(2))`
`B_(P)=B_(3)-(B_(1)+B_(2))`
`=(mu_(0)i)/(2pid)-(mu_(0)i)/(2sqrt(2)pid)=(mu_(0)i)/(4pid)(2-sqrt(2)), o.`
(g)
`d'=d sin.(theta)/(2)`
`B_(1)=(mu_(0)i)/(4pid')(cos. (theta)/(2)+cos 0)`
`=(mu_(0)i)/(4pid sin.(theta)/(2))(1+cos.(theta)/(2))`
`=(mu_(0)i)/(4pid(2sin.(theta)/(4)cos.(theta)/(4)))(1+2cos^(2).(theta)/(4)-1)`
`=(mu_(0)i)/(4pid)cot.(theta)/(4)=B_(2), o.`
`B_(P)=B_(1)+B_(2)=2B_(1)`
`=(mu_(0)i)/(2pid)cot.(theta)/(4), o.`
(h)
`B_(1)=(mu_(0)i)/(2pid)`
`B_(2)=(mu_(0)i)/(4pid)(cos theta+cos theta)`
`cos theta=(l//2)/(sqrt((l//2)^(2)+d^(2))) = l/(sqrt(l^(2)+4d^(2)))`
`B_(1)gtB_(2)`
`B_(1)-B_(2)=1/100B_(1)`
`99/100 B_(1)=B_(2)`
`99/100xx(mu_(0)i)/(2pid)=(mu_(0)i)/(4pid)xx2xxl/(sqrt(l^(2)+4d^(2)))`
`(99)sqrt(l^(2)+4d^(2))=100l`
`(99)^(2)(l^(2)+4d^(2))=(100^(2))l^(2)`
`(99)^(2)4d^(2)=(100^(2)-99^(2))l^(2)`
`(d^(2))/(l^(2))=((199)(1))/((99)^(2)xx4)`
`d/l=0.07`