91影视

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,$\frac{dl}{dt}=k$

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

$\stackrel{鈫�}{B}=\frac{{渭}_{0}NI}{2\mathrm{蟺蝉}}\hat{f}$

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

$蠒={鈭�}_a^{a+w}B.\stackrel{鈫�}{dl}$ 鈥︹�� (1)

Here,$\stackrel{鈫�}{dl}=\left(hds\right)\widehat{蠒}$

Substitute $\frac{{渭}_{0}nl}{2\mathrm{蟺蝉}}\widehat{蠒}$for B and $\left(hds\right)\widehat{蠒}$for into equation (1).

$$\begin{gathered} 蠒={鈭�}_a^{a+w}\left(\frac{{渭}_{0}NI}{2\mathrm{蟺蝉}}\widehat{胃}\right).\left(hds\right)\widehat{蠒} \\ 蠒={鈭�}_a^{a+w}\frac{{渭}_{0}NI}{2蟺}hds \\ 蠒=\frac{{渭}_{0}NIh}{2蟺}{鈭�}_a^{a+w}\frac{1}{s}ds \\ 蠒=\frac{{渭}_{0}NIh}{2蟺}{(\ln s)}_a^{a+w} \end{gathered}$$

Apply the limits,

$$\begin{gathered} 蠒=\frac{{渭}_{0}NIh}{2\mathrm{蟺}}(\ln (a+w)-\ln a) \\ 蠒=\frac{{渭}_{0}NIh}{2\mathrm{蟺}}\ln \left(\frac{a+w}{a}\right) \end{gathered}$$

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

$鈭�E.dl=-\frac{d蠒}{dt}$ 鈥︹�� (2)

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

$鈭�E.dl={渭}_{0}l$ 鈥︹�� (3)

According to the Faraday鈥檚 law of induction,

$鈭�E.dl=-\frac{d}{dt}\left(鈭�B.dl\right)$ 鈥︹�� (4)

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

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

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

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

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

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

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

$$\begin{gathered} l=-\frac{Nh}{2\mathrm{蟺}}\left(\frac{w}{a}\right)k \\ l=-\frac{Nhwk}{2\mathrm{蟺补}} \end{gathered}$$width="112">l=-Nh2蟺wakl=-Nhwk2蟺补

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

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

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

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

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