A long solenoid with $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{turns}\mathbf{/}\mathbf{cm}$and a radius of $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$carries a current of$\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mA}$ . A current of $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{A}$exists in a straight conductor located along the central axis of the solenoid. (a) At what radial distance from the axis will the direction of the resulting magnetic field be at $\mathbf{45}\mathbf{°}$to the axial direction?

(b) What is the magnitude of the magnetic field there?

$\mathrm{n}=10\frac{\mathrm{turns}}{\mathrm{cm}}\mathrm{or}1000\text{\hspace{0.17em}turns/m}$

$\mathrm{R}=7\mathrm{cm}\mathrm{or}0.07\mathrm{m}$

${\mathrm{i}}_{\mathrm{s}}=20\mathrm{mA}\mathrm{or}0.02\mathrm{A}$

${\mathrm{i}}_{\mathrm{w}}=6\mathrm{A}$

The magnetic fieldsdue to solenoid and the long wire are perpendicular to each other.For the net field at${\mathbf{45}}^{\mathbf{0}}$with the axis,we have toequate themagnetic fields to get the radial distance. As the magnitude of the magnetic fieldsare thesame and both are perpendicular to each other, we can calculate the net magnitude of the magnetic field.

Formula:

${\mathrm{B}}_{\mathrm{s}}={\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{s}}\mathrm{n}$

${\mathrm{B}}_{\mathrm{w}}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{w}}}{2\mathrm{\pi d}}$

The net field at a point inside the solenoid is the vector sum of the fields of the solenoid and that of the long straight wire along the central axis of the solenoid.

Thus, $\overrightarrow{\mathrm{B}}=\overrightarrow{{\mathrm{B}}_{\mathrm{s}}}+\overrightarrow{{\mathrm{B}}_{\mathrm{w}}}$

where,$\overrightarrow{{\mathrm{B}}_{\mathrm{s}}}$and$\overrightarrow{{\mathrm{B}}_{\mathrm{w}}}$are the fields due to the solenoid and the wire respectively.

The direction of $\overrightarrow{{\mathrm{B}}_{\mathrm{s}}}$is along the axis of the solenoid, and the direction of $\overrightarrow{{\mathrm{B}}_{\mathrm{w}}}$is perpendicular to it, so the two fields are perpendicular to each other.

For the net field $\overrightarrow{\mathrm{B}}$to be at ${45}^0$with the axis, we must have $\overrightarrow{{\mathrm{B}}_{\mathrm{s}}}=\overrightarrow{{\mathrm{B}}_{\mathrm{w}}}$.

We know the magnetic field due to solenoid is

${\mathrm{B}}_{\mathrm{s}}={\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{s}}\mathrm{n}$

And the magnetic field due to straight wire is

${\mathrm{B}}_{\mathrm{w}}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{w}}}{2\mathrm{\pi d}}$

Therefore, equating the two magnetic fields,

${\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{s}}\mathrm{n}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{w}}}{2\mathrm{\pi d}}$

Then the radial distance $\mathrm{d}$is

$\begin{array}{c}\mathrm{d}=\frac{{\mathrm{i}}_{\mathrm{w}}}{2{\mathrm{\pi i}}_{\mathrm{s}}\mathrm{n}}\\ =\frac{6\text{\hspace{0.17em}A}}{2\mathrm{\pi }\times 0.02\text{\hspace{0.17em}A}\times 1000\text{\hspace{0.17em}turns/m}}\\ =0.047\mathrm{m}\end{array}$

The radial distance from the axis at which the direction of the resulting magnetic field will be at $45°$to the axial direction is $\mathrm{d}=0.047\mathrm{m}$

The magnetic field strength is given by the equation

$\mathrm{B}=\sqrt{{\mathrm{B}}_{\mathrm{s}}^2+{\mathrm{B}}_{\mathrm{w}}^2+2{\mathrm{B}}_{\mathrm{s}}{\mathrm{B}}_{\mathrm{w}}\mathrm{cos\theta }}$

Since both the magnetic fields are perpendicular to each other and their magnitudes arethesame, i.e.,

${\mathrm{B}}_{\mathrm{s}}={\mathrm{B}}_{\mathrm{w}}$and $\mathrm{\theta }=90°$

So, $\mathrm{B}=\sqrt{{\mathrm{B}}_{\mathrm{s}}^2+{\mathrm{B}}_{\mathrm{s}}^2}$

$\mathrm{B}=\sqrt{2}{\mathrm{B}}_{\mathrm{s}}$

$\begin{array}{c}\mathrm{B}=\sqrt{2}{\mathrm{\mu }}_0{\mathrm{i}}_{\mathrm{s}}\mathrm{n}\\ =\sqrt{2}\times 4\mathrm{\pi }\times {10}^{-7}{\text{\hspace{0.17em}NA}}^{\text{-2}}\times 0.02\text{\hspace{0.17em}A}\times 1000\text{\hspace{0.17em}turns/m}\\ =3.55\times {10}^{-5}\text{T}\end{array}$

The magnitude of the magnetic field is $\mathrm{B}=3.55\times {10}^{-5}\mathrm{T}$.