A long solenoid with 9.95 turns/cm and a radius of 7.81 cm carries a current of 29.6 mA. A...

Question:

A long solenoid with 9.95 turns/cm and a radius of 7.81 cm carries a current of 29.6 mA. A current of 9.45 A exists in a straight conductor located along the central axis of the solenoid.

At what radial distance from the axis in centimeters will the direction of the resulting magnetic field be at 39.3° to the axial direction?

Magnetic Field & Current:

A solenoid is a form of inductor which is a coiled current-carrying conductor wire form as a long cylindrical form. The solenoid is however used as an electromagnet which is used to produce controlled magnetic fields.

1. N = Number of turns per length = 9.95 turns/cm = 995 turns/ m
2. R = Radius of the solenoid = 7.81 cm = 0.0781 m
3. I = Current carried by the solenoid = 29.6 mA = 0.0296 A
4. Is = Current carried by the straight conductor = 9.45 A
5. {eq}\theta {/eq} = 39.3°

The mathematical expression formed by the given conditions is,

{eq}\tan\theta = \dfrac{B_w}{B_s}\\ \textrm{where,}\\ B_s = \textrm{magnetic field due to the solenoid} = \mu_0 nI_s\\ B_w = \textrm{magnetic field due to the wire} = \dfrac{\mu_0 I_w}{2\pi\; r}\\ {/eq}

Applying the values we get,

{eq}\tan\theta = \dfrac{\mu_0I_w}{2\pi\;r(\mu_0nI_s)}\\ \textrm{For determining,}r \\ r = \dfrac{I_w}{2\pi\; n I_s\tan\theta}\\ r = \dfrac{9.45}{6.28(995)(0.0296)\tan39.3°}\\ r = 0.0624\;\mathrm{m} {/eq} 