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

Explanation

(a)
Magnetic field due to straight wires=0.
Magnetic field due to a semicircular wire
`B_("semicircular")=(mu_0I)/(2R) (theta/(2pi))=(muI)/(2R) (pi/(2pi))`
or `B_("semicircular")=(mu_0I)/(4R)`
The direction can be verified from the right hadn rule that
will be down ward into the plane of the page.
(b)
Magnetic field due to straight wire 1-2 will be zero. as in its
direction is passing through centre of the circular wire.
Magnetic field due to circular part
`B_("circular")=(mu_0I)/(2R) (theta/(2pi))=(mu_0I)/(2R) (((3//2)pi)/(2pi))`
`B_("circular")=(3mu_0I)/(8R) ox`
Magnetic field due to semi-infinite wire 3-4,
`B_("straight")=(mu_0I)/(4piR) ox`
Net magnetic field at O
`B_0=(3mu_0I)/(8R)+(mu_0I)/(4piR)`
`B_0=(mu_0I)/(4piR)(3/2+1/(pi))ox`
(c)
Magnetic field at O
`vecB_0=vecB_("circle")+vecB_("straight") =(mu_0I)/(2R)odot +(mu_0I)/(2piR)odot`
` implies B=(mu_0I)/(2R)(1+1/(pi))odot`
(d) The magnetic field due to straight wires
will be zero. Magnetic field due to circular
part
`B_("circular")=(mu_0I)/(2R)(theta/(2pi))=(mu_0I)/(2R)(((3//2)pi)/(2pi))=(3mu_0I)/(8R)ox`
(e) `vecB_R=((mu_0I)/(4piR)+(mu_0I)/(4R)+(mu_0I)/(4piR))(-hatk)=(mu_0I)/(2R)(1/pi+1/2) hatk`