Calculate the magnetic field at point O in each of the following cases:

Explanation

Here straight and circular shape wires are joined, determine magnetic field for each shape and find resultant magnetic field. It should be kept in mind that magnetic field due to stright wire at any point along its length is zero.
(a)
Since point `O` is along the length of straight wires `1` and `2` hence `B_(1)=B_(3)=O`
`B_(O)=B_(2)=(mu_(0)i)/(4R), ox`
(b)
Point `O` is the foot of long wires `(1)` and `(2)`
`B_(1)=B_(3)=(mu_(0)i)/(4piR), o.`
Due to semi-circle `(2)`
`B_(2)=(mu_(0)i)/(4R), o.`
`B_(O)=B_(1)+B_(3)+B_(2)`
`=(mu_(0)i)/(2piR)+(mu_(0)i)/(4R)`
`=(mu_(0)i)/(4R)(2/(pi)+1),o.`
(c)
At `O`:
`B_(1)=(mu_(0)i)/(4piR),ox`
`B_(2)=(mu_(0)i)/(4R),ox`
`B_(3)=0`
`B_(O)=B_(1)+B_(2)=(mu_(0)i)/(4piR)+(mu_(0)i)/(4R)`
`(mu_(0)i)/(4R)(1/(pi)+1) ox`
(d)
Magnetic field at `O` due to straight wires is zero.
`B_(1)=(mu_(0)i)/(4R_(1)), ox`
`B_(2)=(mu_(0)i)/(4R_(2)), ox`
`B_(O)=B_(1)+B_(2)=(mu_(0)i)/4(1/(R_(1))+1/(R_(2))),ox`
(e)
Magnetic field at `O` due to straight wires is zero.
`B_(1)=(mu_(0)i)/(4R_(1)), o.`
`B_(2)=(mu_(0)i)/(4R_(2)), ox`
`B_(1)gtB_(2)`
`B_(O)=B_(1)-B_(2)=(mu_(0)i)/4(1/(R_(1))-1/(R_(2))),o.`
(f)
Due to straight wire

`B_(1)=((mu_(0)i)/(2piR)), o.`
Due to circular wire
`B_(2)=((mu_(0)i)/(2R)), o.`
`B_(O)=B_(1)+B_(2)=(mu_(0)i)/(2R)(1/pi+1),o.`
(g)
`B_(1)=((mu_(0)i)/(4piR)), o.`
`B_(2)=((mu_(0)i)/(2R)), ox`
`B_(3)=((mu_(0)i)/(4piR)), o.`
`B_(2)gt(B_(1)+B_(3))`
`B_(O)=B_(2)-(B_(1)+B_(3))`
`=(mu_(0)i)/(2R)-(mu_(0)i)/(2piR)(1/pi-1),ox`
(h)
`B_(1)=((mu_(0)i)/(4piR)), ox`
`B_(2)=((mu_(0)i)/(2R)).3/4,ox`
`B_(3)=0`
`B_(O)=B_(1)+B_(2)`
`=(mu_(0)i)/(4R)(1/pi+3/2),ox`
(i)
`B_(1)=(mu_(0)i)/(2a).(theta)/(2pi), o.`
`B_(2)=(mu_(0)i)/(2b).(theta)/(2pi), ox`
`B_(1)gtB_(2)`
`B_(O)=B_(1)-B_(2)`
`=(mu_(0)i theta)/(4pi)(1/a-1/b), o.`
(j)
`d=R cos (theta)/(2)`
`B_(1)=(mu_(0)i)/(4pid)(sin.(theta)/(2) + sin. (theta)/(2))`
`=(mu_(0)i)/(4pi.Rcos. (theta)/(2)).2sin.(theta)/(2)`
`=(mu_(0)i)/(2piR)tan. (theta)/(2),ox`
`B_(2)=(mu_(0)i)/(2R)((2pi-theta))/(2pi),ox`
`B_(O)=B_(1)+B_(2)`
`=(mu_(0)i)/(2piR)(tan.(theta)/(2)+pi-theta/2),ox`
(k)
`B_(1)=(mu_(0)i)/(2R_(1)). ((2pi-theta))/(2pi), ox`
`B_(2)=(mu_(0)i)/(2R_(2)).(theta)/(2pi),ox`
`B_(O)=B_(1)+B_(2)`
`=(mu_(0)i)/(4piR_(1))(2 pi -theta)+(mu_(0)i)/(4piR_(2)) theta, ox`
(l)


`B_(1)=(mu_(0)i)/(2a).3/4=(3mu_(0)i)/(8a),ox`
`B_(2)=(mu_(0)i)/(4pib) (cos90^(@)+cos45^(@))`
`=(mu_(0)i)/(4sqrt(2)pib)=B_(3),ox`
`B_(O)=B_(1)+(B_(2)+B_(3))`
`=(3mu_(0)i)/(8a)+(mu_(0)i)/(2sqrt(2)pib), ox`
(m)
`d=asin30^(@)=a//2`
`B_(1)=(mu_(0)i)/(4pid) (cos30^(@)+cos30^(@))`
`=(mu_(0)i)/(4pi.a/2).2.(sqrt(3))/2=(sqrt(3)mu_(0)i)/(2pia),ox`
`B_(2)=(mu_(0)i)/(2a).(120^(@))/(360^(@))=(mu_(0)i)/(6a),o.`
`B_(1)gtB_(2)`
`B_(O)=B_(1)-B_(2)`
`=(mu_(0)i)/(2a)((sqrt(3))/(pi)-1/3), ox`