Find the magnetic induction of the field at the point `O` at a loop with current `I`, whose shape is illustrated in figure
(a)In figure `a` the radii `a` and `b`, as well as the angle `varphi` are known
(b)In figure `b`,the raidus `a` and the side `b` are known.
(c)A current `I=5.0 A` flows along a thin wire shaped as shown in figure.The radius of a curved part of the wire is equal to `R=120 mm`,the angle `2varphi=90^(@)`.Find the magnetic induction of the field at the point `O`.

Answer

A, B, C, D

Explanation

`B_("due to st.part AB")=0=B_("due to st.part CD")`
`B_("due to curved part AD")=((2pi-phi)/(2pi))((mu_(0)i)/(2a))` Into the plane of paper.
`B_("due to curved part BC")=(phi)/(2pi)((mu_(0)i)/(2b))` Into the plane of paper.
`B_(Net)=(mu_(0)i)/(4pi)[(2pi-phi)/a+phi/b]` Into the plane of paper.
(b)`B_("due to BC")=B_("due to EA")=0`
`B_("due to curved part AB")=((3pi)/2)/(2pi)(mu_(0)I)/(2a) rArr =3/8(mu_(0)l)/a` Into the plane of paper
`B_("due to CD")=(mu_(0)i)/(4pib)[cos 90^(@)+cos 45^(@)]rArr=(mu_(0)I)/(4sqrt2pib)` Into the paper
`B_("due to DE")=(mu_(0)I)/(4sqrt2pib)` Into the plane of paper
`vecB_(Net)=(mu_(0)I)/(4pi)[(3pi)/(2a)-sqrt2/b]` Into the plane of paper.
(c) `B=B_("due to st.part")+B_("due to curved part")` both into the plane of paper
`=((2pi-2phi)/(2pi))(mu_(0)i)/(2R)+(mu_(0)i)/(4pid)[sin phi+sin phi]`
`=((2pi-2phi)/(2pi))(mu_(0)i)/(2R)+(mu_(0)i)/(4pi R cos phi)[2 sin phi]`
`(mu_(0)i)/(2piR)[pi-phi+tanphi]=28muT`