A long wire carries a current of 18 A kept along the axis of a long solenoid of radius 1 cm. The field due to the solenoid is `8 xx10 ^(-3)T.` The magnitude of the resultant field at a point 0.6 mm from the solenoid axis is
(Assume `mu_(0) = 4pi xx 10 ^(-7)Tm//A)`

Options

A. `6xx 10^(-3) T`
B. ` 6xx 10^(-4)T`
C. `2 sqrt7 xx 10 ^(-3) T`
D. ` 10 xx 10^(-3) T`

Answer

D

Explanation

Given,
Current in a long wire, `I = 18A`
Radius of solenoid, `R =1 cm`
Field due to solenoid, `B _(1) =8 xx 10 ^(-3) T`
`r = 0.6m`
`mu _(0) = 4pi xx 10 ^(-7) Tm//A`
Magnetic field due to current carrying wire is given by,
`B _(2) = (mu_(0))/(2pi) xx (I)/(r)`
`= (4pi xx 10 ^(-7))/(2pi) xx (18)/( 0.6 xx 10 ^(-3))`
`= 6xx 10 ^(-3)T`
Resultant fiedl magnitude, `B = sqrt(B_(1) ^(2) + B_(2) ^(2))`
`= sqrt((8xx 10^(-3)) ^(2) + (6xx 10 ^(-3))^(2)) `
`=sqrt(64 xx 10 ^(-6) + 36+10 ^(-6))`
`= sqrt(100xx 10 ^(-6))`
`therefore B = 10 xx 10 ^(-3) T`