Chapter 5: Q5.18P (page 240)

Show that the magnetic field of an infinite solenoid runs parallel to the axis, regardless of the cross-sectional shape of the coil,as long as that shape is constant along the length of the solenoid. What is the magnitude of the field, inside and outside of such a coil? Show that the toroid field (Eq. 5.60) reduces to the solenoid field, when the radius of the donut is so large that a segment can be considered essentially straight.

Short Answer

Expert verified

The magnetic field runs parallel to the axis of solenoid regardless of the shape, as along the shape is constant along the length of the solenoid.

Step by step solution

Define function

According to Biot-Savart’s law, write the expression of magnetic field at point at distance $r$ .

$B = \frac{μ_{0} I}{4 π} \int \frac{dI \times r}{r^{3}}$ …… (1)

Here, $μ_{0}$ is the permeability for free space, $I$ is the current, $r$ is the distance and $d l$ is the element.

Determine figure

Consider the elements $d l_{1}$ and $d l_{2}$ at points $P (x^{'}, y^{'}, z^{'})$ and $P^{'} (x^{'}, y^{'}, - z^{'})$ respectively.

The points $P$ and $P^{'}$ lie symmetrically with respect to x-y plane. Also assume a point $M (0, y, 0)$ located on y-axis.

Determine magnetic field

Write the expression for the magnetic field due to the elements.

$d B = \frac{μ_{0} I}{4 π} (\frac{d I_{1} \times r_{1}}{r_{1}^{3}} + \frac{d I_{2} \times r_{2}}{r_{2}^{3}})$ …… (2)

Here, $r_{1}$ and $r_{2}$ are the position vectors of point $P$ and $P'$ from $M$ respectively.

From the above figure,

Write the expression for position vector $r_{1}$ .

$r_{1} = r_{M} - r_{P}$

Substitute $y \hat{y}$ for $r_{M}$ and $x^{'} \hat{x} + y^{'} \hat{y} + z^{'} \hat{z}$ for $r_{P}$ in above equation.

$\begin{matrix} r_{1} = r_{M} - r_{P} \\ = y \hat{y} - (x^{'} \hat{x} + y^{'} \hat{y} + z^{'} \hat{z}) \\ = - x^{'} \hat{x} + (y - y^{'}) \hat{y} - z^{'} \hat{z} \end{matrix}$

Write the magnitude of $r_{1}$ .

$\begin{matrix} r_{1} = \sqrt{{(- x^{'})}^{2} + {(y - y^{'})}^{2} + {(- z^{'})}^{2}} \\ = \sqrt{{x^{'}}^{2} + {(y - y^{'})}^{2} + {z^{'}}^{2}} \end{matrix}$

From the above figure,

Write the expression for position vector $r_{2}$ .

$r_{2} = r_{M} - r_{P^{'}}$

Substitute $y \hat{y}$ for $r_{M}$ and $x^{'} \hat{x} + y^{'} \hat{y} - z^{'} \hat{z}$ for $r_{P}$ in above equation.

$\begin{matrix} r_{2} = r_{M} - r_{P^{'}} \\ = y \hat{y} - (x^{'} \hat{x} + y^{'} \hat{y} - z^{'} \hat{z}) \\ = - x^{'} \hat{x} + (y - y^{'}) \hat{y} + z^{'} \hat{z} \end{matrix}$

Write the magnitude of $r_{1}$ .

$\begin{matrix} r_{2} = \sqrt{{(- x^{'})}^{2} + {(y - y^{'})}^{2} + {(- z^{'})}^{2}} \\ = \sqrt{{x^{'}}^{2} + {(y - y^{'})}^{2} + {z^{'}}^{2}} \end{matrix}$

Thus, the magnitude of $r_{1}$ and $r_{2}$ are equal.

$r_{1} = r_{2} = r$

Write the expression for $d l_{1}$ and $d l_{2}$ .

$\begin{array}{l} d I_{1} = d x^{'} \hat{x} + d y^{'} \hat{y} \\ d I_{2} = d x^{'} \hat{x} + d y^{'} \hat{y} \end{array}$

Thus, the two elements are equal.

$d I_{1} = d I_{2} = d I$

Substitute $d l$ for $d I_{1}, d I_{2}$ and $r$ for $r_{1}$ and $r_{2}$ in equation (2)

$\begin{matrix} d B = \frac{μ_{0} I}{4 π} (\frac{d I_{1} \times r_{1}}{r_{1}^{3}} + \frac{d I_{2} \times r_{2}}{r_{2}^{3}}) \\ = \frac{μ_{0} I}{4 π} (\frac{d I \times (r_{1} + r_{2})}{r^{2}}) \end{matrix}$ …… (3)

Determine magnetic field

As $d l_{1}$ and $(r_{1} + r_{2})$ are in the same x-y plane, $d B ‖ d I_{1} \times (r_{1} + r_{2})$ is along with z axis which is perpendicular to x-y plane.

Substitute $(d x^{'} \hat{x} + d y^{'} \hat{y})$ for $d l$ , $- x^{'} \hat{x} + (y - y^{'}) \hat{y} - z^{'} \hat{z}$ for $r_{1}$ , $- x^{'} \hat{x} + (y - y^{'}) \hat{y} + z^{'} \hat{z}$ for $r_{2}$ ,

$\sqrt{{x^{'}}^{2} + {(y - y^{'})}^{2} + {z^{'}}^{2}}$ for $r$ in equation (3).

$\begin{matrix} d B = \frac{μ_{0} I}{4 π} (\frac{d I \times (r_{1} + r_{2})}{r^{2}}) \\ = \frac{μ_{0} I}{4 π} (\frac{(d x^{'} \hat{x} + d y^{'} \hat{y}) \times (- x^{'} \hat{x} + (y - y^{'}) \hat{y} - z^{'} \hat{z}) + (- x^{'} \hat{x} + (y - y^{'}) \hat{y} + z^{'} \hat{z})}{{(\sqrt{{x^{'}}^{2} + {(y - y^{'})}^{2} + {z^{'}}^{2}})}^{3}}) \\ = \frac{μ_{0} I}{4 π} (\frac{(d x^{'} \hat{x} + d y^{'} \hat{y}) \times (- 2 x^{'} \hat{x} + 2 (y - y^{'})) \hat{y}}{(\sqrt{{x^{'}}^{2} + {(y - y^{'})}^{2} + {z^{'}}^{2}})}) \\ = \frac{μ_{0} I}{4 π} (\frac{2 (y - y^{'}) d x^{'} + 2 x^{'} d y^{'}}{{({(x^{'})}^{2} + {(y - y^{'})}^{2} + {(z^{'})}^{2})}^{\frac{3}{2}}}) \hat{z} \end{matrix}$

From above it is clear that, the filed is running parallel to the axis of solenoid that is along z axis.

Therefore, the magnetic field runs parallel to the axis of solenoid regardless of the shape, as along the shape is constant along the length of the solenoid.