Question

A long solenoid with $ 10 $ turns/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 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
$I_{1}=20.0 mA=20\times10^{-3}\,A$
$B_{1}$= magnetic field due to solenoid
$B_{2}$= magnetic field due to wire
For wire $I_{2}=6\,A$
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, $B_{1}=B_{2}$
$\Rightarrow \mu_{0} \, nI_{1}=\frac{\mu_{0}I_{2}}{2\pi r}$
$\Rightarrow r=\frac{I_{2}}{I_{1}\times n2\,\pi}=\frac{6}{20\times10^{-3}\times10\times2\times3.14}$
$=\frac{6\times10^{2}}{20\times2\times3.14}=\frac{60}{4\times3.14}$
$=4.77$
$=4.8\,cm$ (approx)