Question

A toroidal coil has a rectangular cross section, with inner radius a , outer radius $\mathbf{a}\mathbf{+}\mathbf{w}$, and height h . It carries a total of N tightly wound turns, and the current is increasing at a constant rate $\mathbf{(}\mathbf{dl}\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{k}\mathbf{)}$. If w and h are both much less than a , find the electric field at a point z above the center of the toroid. [Hint: Exploit the analogy between Faraday fields and magnetostatic fields, and refer to Ex. 5.6.]

Short answer

The electric field at point above the centre of the toroid is$-\frac{\mu }{4\mathrm{\pi }}\frac{Nhwka}{{\left(a^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}\widehat{z.}$

Step-by-step solution

Write the given data from the question.

The inner radius of the toroidal is a.

The outer radius of toroidal is $a+w.$

The height of toroidal is h.

The number of the turns is N.

Current increasing constant rate,$\frac{dl}{dt}=k$

Determine the electric field at a point z above the centre of the toroid.

The expression for the magnetic field inside the toroid is given by,

$\overrightarrow{B}=\frac{{\mu }_{0}NI}{2\mathrm{\pi s}}\hat{f}$

The flux around the toroid is calculated by the integral of product of magnetic field and area.

$\varphi ={\int }_a^{a+w}B.\overrightarrow{dl}$ …… (1)

Here,$\overrightarrow{dl}=\left(hds\right)\hat{\varphi }$

Substitute $\frac{{\mu }_{0}nl}{2\mathrm{\pi s}}\hat{\varphi }$for B and $\left(hds\right)\hat{\varphi }$for into equation (1).

$$\begin{gathered} \varphi ={\int }_a^{a+w}\left(\frac{{\mu }_{0}NI}{2\mathrm{\pi s}}\hat{\theta }\right).\left(hds\right)\hat{\varphi } \\ \varphi ={\int }_a^{a+w}\frac{{\mu }_{0}NI}{2\pi }hds \\ \varphi =\frac{{\mu }_{0}NIh}{2\pi }{\int }_a^{a+w}\frac{1}{s}ds \\ \varphi =\frac{{\mu }_{0}NIh}{2\pi }{(\ln s)}_a^{a+w} \end{gathered}$$

Apply the limits,

$$\begin{gathered} \varphi =\frac{{\mu }_{0}NIh}{2\mathrm{\pi }}(\ln (a+w)-\ln a) \\ \varphi =\frac{{\mu }_{0}NIh}{2\mathrm{\pi }}\ln \left(\frac{a+w}{a}\right) \end{gathered}$$

According to the Faraday’s law, the electric field around the closed surfaces is given as,

$\oint E.dl=-\frac{d\varphi }{dt}$ …… (2)

For the magnetostatics, the integral of magnetic field is given as,

$\oint E.dl={\mu }_{0}l$ …… (3)

According to the Faraday’s law of induction,

$\oint E.dl=-\frac{d}{dt}\left(\oint B.dl\right)$ …… (4)

From the equations (2), (3), and (4).

$$\begin{gathered} {\mu }_{0}l=-\frac{d\varphi }{dt} \\ l=-\frac{1}{{\mu }_0}\frac{d\varphi }{dt} \end{gathered}$$

Substitute $\frac{{\mu }_{0}Nlh}{2\mathrm{\pi }}\ln \left(\frac{a+w}{a}\right)$for $\varphi$into above equation.

$$\begin{gathered} l=-\frac{1}{{\mu }_0}\frac{d}{dt}\left[\frac{{\mu }_{0}Nlh}{2\mathrm{\pi }}in\left(\frac{a+w}{a}\right)\right] \\ l=-\frac{{\mu }_{0}Nh}{2{\mathrm{\pi \mu }}_0\mathrm{s}}\ln \left(1+\frac{w}{a}\right)\frac{dl}{dt} \end{gathered}$$

By approximation,$\ln \left(1+\frac{w}{a}\right)\approx \frac{w}{a}$

$l=\frac{Nh}{2\mathrm{\pi }}\left(\frac{w}{a}\right)\frac{dl}{dt}$

Substitute K for $\frac{dl}{dt}$into above equation.

$$\begin{gathered} l=-\frac{Nh}{2\mathrm{\pi }}\left(\frac{w}{a}\right)k \\ l=-\frac{Nhwk}{2\mathrm{\pi a}} \end{gathered}$$width="112">l=-Nh2πwakl=-Nhwk2πa

The electric and magnetic field will be at the point z.

$$\begin{gathered} E=B \\ E=\frac{{\mu }_{0}l}{2}\frac{a^2}{{\left(a^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}\hat{z} \end{gathered}$$

Substitute$\frac{Nhwk}{2\mathrm{\pi a}}$ for$l$ into above equation.

$$\begin{gathered} E=\frac{{\mu }_0}{2}\left(-\frac{Nhwk}{2\mathrm{\pi a}}\right)\frac{a^2}{{\left(a^2+z^2\right)}^{{\displaystyle \frac{3}{2}}}}\hat{z} \\ E=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{Nhwka}{{\left(a^2+z^2\right)}^{\frac{3}{2}}}\hat{z} \end{gathered}$$

Hence the electric field at point z above the centre of the toroid is $-\frac{{\mu }_0}{4\mathrm{\pi }}\frac{Nhwka}{{\left(a^2+z^2\right)}^{\frac{3}{2}}}\hat{z}$