Question

A long solenoid with $10$ turns/cm and a radius of $7.0cm$ carries a current of $20.0mA$ . A current of $6.0A$ exists in a straight conductor located along the central axis of the solenoid. At what radial distance from the axis will the direction of the magnetic field be at $45^{\circ }$ to the axial direction?

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

Answer 1

Answer: (A)

Given, for solenoid
$n=10$ turn/cm
$I_1=20.0mA=20\times 10^{-3}A$
$B_1$= magnetic field due to solenoid
$B_2$= magnetic field due to wire
For wire $I_2=6A$
Now, according to question if resultant should be at $45^{\circ }$ then the two field $B_1$ and $B_1$ must be equal to each other
So, $B_1=B_2$
$\Rightarrow {\mu }_{0}n{I}_1=\frac{{\mu }_0{I}_2}{2\pi r}$
$\Rightarrow r=\frac{I_2}{I_1\times n2\pi }=\frac{6}{20\times 10^{-3}\times 10\times 2\times 3.14}$
$=\frac{6\times 10^2}{20\times 2\times 3.14}=\frac{60}{4\times 3.14}$
$=4.77$
$=4.8cm$ (approx)