91影视

The side of the square loop is$\mathrm{a}$

The distance between the long wires is$3\mathrm{a}$

The current is the square loop is$\mathrm{l}$

The increasing rate of the current $\raisebox{1ex}{$\mathrm{dl}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.=\mathrm{k}$,

Consider the diagram of the system as,

The expression for the magnetic field of one wire is given by,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{渭}}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{蟺蝉}}$

The magnetic flus due to one wire is given by,

${\mathrm{蠒}}_1鈭�{鈭�}_{\mathrm{胃}}^{2\mathrm{a}}鈥�\mathrm{B}脳\mathrm{da}$

Substitute $\frac{{\mathrm{渭}}_0\mathrm{f}}{2\mathrm{蟺蝉}}$for$\mathrm{B}$and $\mathrm{ads}$for $\mathrm{da}$into the above equation.

${\mathrm{蠒}}_1={鈭�}_{\mathrm{a}}^{2\mathrm{a}}鈥�\frac{{\mathrm{渭}}_0\mathrm{l}}{2\mathrm{蟺蝉}}脳\mathrm{ads}$

${\mathrm{蠒}}_1=\frac{{\mathrm{渭}}_0\mathrm{la}}{2\mathrm{蟺}}{鈭�}_{\mathrm{a}}^{2\mathrm{a}}鈥�\frac{1}{\mathrm{s}}\mathrm{ds}$

${\mathrm{蠒}}_1=\frac{{\mathrm{渭}}_0\mathrm{la}}{2\mathrm{蟺}}(\ln 鈦�\mathrm{s}{)}_{\mathrm{a}}^{2\mathrm{a}}$

${\mathrm{蠒}}_1=\frac{{\mathrm{渭}}_0\mathrm{la}}{2\mathrm{蟺}}(\ln 鈦�2)$

The magnetic flue due to two wire is given by,

$\mathrm{蠒}=2{\mathrm{蠒}}_1$

$\mathrm{蠒}=2\frac{{\mathrm{渭}}_0\mathrm{la}}{2\mathrm{蟺}}\ln 鈦�\left(2\right)$

$\mathrm{蠒}=\frac{{\mathrm{渭}}_0\mathrm{la}}{\mathrm{蟺}}\ln 鈦�\left(2\right)$

The induced emf is given by,

$\mathrm{蔚}=鈭�\frac{\mathrm{诲蠒}}{\mathrm{dt}}$

Substitute ${\mathrm{渭}}_{\mathrm{蟺}}{\mathrm{渭}}_0\ln 鈦�\left(2\right)$for $\mathrm{蠒}$into above equation

$\mathrm{蔚}=鈭�\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{{\mathrm{渭}}_0\mathrm{la}}{\mathrm{蟺}}\ln 鈦�\left(2\right)\right)$

$\mathrm{未}=鈭�\frac{{\mathrm{渭}}_0/\mathrm{a}}{\mathrm{蟺}}\ln 鈦�\left(2\right)\frac{\mathrm{di}}{\mathrm{dt}}$

Substitute $\mathrm{k}$for $\raisebox{1ex}{$\mathrm{dl}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{dt}$}\right.$into the above equation.

$\mathrm{蔚}=鈭�\frac{{\mathrm{渭}}_0\mathrm{a}}{\mathrm{蟺}}\ln 鈦�\left(2\right)\mathrm{k}$

$\mathrm{蔚}=鈭�\frac{{\mathrm{渭}}_0\mathrm{ka}}{\mathrm{蟺}}\ln 鈦�\left(2\right)$

Therefore, the induced emf is $鈭�\frac{{\mathrm{a}}_0\mathrm{ka}}{\mathrm{蟺}}\ln 鈦�\left(2\right)$.

The net flux in inward and increasing. According to Len鈥檚 law, the emf should oppose the increment in the flux. Therefore, the magnetic flux should be outward, and the current flows in a counter clockwise direction in the big loop.

Hence the induced emf is $鈭�\frac{{\mathrm{a}}_0\mathrm{ka}}{\mathrm{蟺}}\ln 鈦�\left(2\right)$and direction of the current in big loop is counter clockwise.