A long wire bent as shown in the figure carries current `I=10A`. If the radius of the semicircular portion is 1 m, the magnetic induction `("in "muT)` at the centre C is

Options

A. `pi ^(2) + 4`
B. `sqrt ( pi ^(2) + 4)`
C. `pi ^(2) - 4`
D. `sqrt ( pi ^(2) - 4)`

Answer

B

Explanation

Using right hand thumb rule, one can determine that both the straight wires will have same direction and magnitude of magnetic field at centre of circle. Similarly magnetic field due to circular coil will be perpendicular to the magnetic field due to straight wires.
Total magnetic field induction at C is
Magnetic field due to semi circular loop,
`B _(1) = (1)/(2) (( mu _(0)I )/( 2a )) - hati `
Magnetic field due to two semi infinite wires,
`B _(2) = 2 xx ((1)/(2) ( mu _(0) I )/( 2pi a )) hatk`
Resultant magnetic field will be ,
`B = sqrt (B _(1) ^(2) + B _(2) ^(2))`
`= ( mu _(0) I )/(2a) sqrt (((1)/(2)) ^(2) + ((1)/(pi))^(2))`
`= ( mu _(0) I )/( 4 pia) sqrt (pi ^(2) + 4)`
`a =1, i = 10`
`= ( 4pi xx 10 ^(-7) xx 10)/( 4pi xx 1) sqrt ( pi ^(2) +4)`
`= sqrt ( pi ^(2) +4) mu T`