91影视

Let S be the surface bounded by the loop (P) at time t .

Let S' a surface bounded by the loop in its new position (P') at time t+dt .

Let R is the "ribbon" joining P and P' .

Write the formula ofohm鈥檚 law using faraday鈥檚 law.

$\mathbf{J}\mathbf{=}\mathbf{蟽}\left(E+v脳B\right)$ 鈥︹�� (1)

Here, $\mathit{蟽}$ is charge density, Eis electrical field andB is magnetic field.

Write the formula of invoke stokes鈥� theorem.

$\mathbf{鈭�}\mathbf{B}\mathbf{路}\mathbf{da}\mathbf{=}\mathbf{0}$ 鈥︹�� (2)

Here,B is magnetic field and da is the radius of the circle.

According to ohm鈥檚 law

Substitute 0 for J and $鈭�$for $蟽$into equation (1).

$\left(\mathrm{E}+\mathrm{v}脳\mathrm{B}\right)=0$

Taking curl on both sides then

$鈭�脳\mathrm{E}+鈭�脳\left(\mathrm{v}脳\mathrm{B}\right)=0$

From faraday鈥檚 law

$鈭�脳E=-\frac{鈭�B}{鈭�t}$

Then:

$$\begin{gathered} -\frac{鈭�B}{鈭�t}+鈭�脳\left(鈭�脳B\right)=0 \\ \frac{鈭�B}{鈭�t}=鈭�脳\left(鈭�脳B\right) \end{gathered}$$

Therefore, the value to prove that $\frac{鈭�B}{鈭�t}=鈭�脳\left(鈭�脳B\right)$.

As we know that for any closed surface.

$鈭�脳\mathrm{B}=0$

Determine invoke stokes鈥� theorem.

Substitute ${鈭�}_{S\text{'}}B\left(t+dt\right)路da-{鈭�}_{R}B\left(t+dt\right)路da-{鈭�}_{S}B\left(t\right).da\mathrm{for}鈭�\mathrm{B}$into equation (2).

${鈭�}_{S\text{'}}B\left(t+dt\right)路da-{鈭�}_{R}B\left(t+dt\right)路da-{鈭�}_{S}B\left(t\right).da=0$

Here:

$$\begin{gathered} d蠒={鈭�}_{S\text{'}}B\left(t+dt\right)路da-{鈭�}_{S}B\left(t\right)路da-{鈭�}_{R}B\left(t+dt\right)路da \\ ={鈭�}_S\left[B\left(t+dt\right)路da-B\left(t\right)\right]路da-{鈭�}_{R}B\left(t+dt\right)路da \\ =\left\{{鈭�}_S\left(\frac{鈭�B}{dt}路dt\right)dt-{鈭�}_{R}B\left(t+dt\right)路da\right\} \\ ={鈭�}_S\left(\frac{鈭�B}{dt}路dt\right)dt-{鈭�}_{R}B\left(t+dt\right)路\left(\mathrm{dI}脳\mathrm{v}\right)dt \end{gathered}$$

Solve further as

Then:

$$\begin{gathered} d蠒=dt{鈭�}_S\frac{鈭�B}{鈭�t}路da-dt{鈭�}_{S}B路\left(\mathrm{dI}脳\mathrm{v}\right) \\ d蠒=dt\left\{{鈭�}_S\left(\frac{鈭�B}{鈭�t}\right)路da-{鈭�}_S鈭�脳\left(\mathrm{v}脳\mathrm{B}\right)路da\right\} \\ \frac{d蠒}{dt}={鈭�}_S\left[\frac{鈭�B}{鈭�t}-鈭�脳\left(v脳B\right)\right]路da=0 \end{gathered}$$