A long solenoid with 10 turn / cm and a radius of 7.0 cm carries a current of 20.0 mA .A current of 6.0 A exists in a straight conductor loacted along the central axis of the solenoid at what radial distance from the axis will the direction of the magneitic field be at `45^(@)` to the axial direction

Options

A. 4.8 cm
B. 8.1 cm
C. 9.9 c
D. 10.6 cm

Answer

A

Explanation

Given for solenoid
n=10 truns /cm
`I_(1)=20.0 mA=20xx10^(-3) A`
`B_(1)` = magnetic field due to solenoid
`B_(2)` = magnetic field due to solenoid ltDbrgt `B_(2)` = magnetic field due to wire
for wire `l_(2)=6A`
Now according to question if resultant should be at `45^(@)` then the two field `B_(1)` and `B_(2)` must be equal to each other
so `B_(1)=B_(2) rarr mu_(0)nl_(1)=(m_(0)l_(2))/(2pir)`
`rarr` radius `r=(l_(2))/(l_(1)xxn2pi)`
`=(6)/(20xx10^(-3)xx102xx3.14)`
`=(6xx10^(2))/(20xx2xx3.14)=(60)/(4xx3.14)`
=4.77 =4.8 cm (approx)