91影视

The reservoir empties only when switched is closed after a lengthy moment$t>0$, and electrons flow from the gate electrode to the terminal in a circular fashion, with the water passing from right to left in the bottom cable.

To estimate the direction of the induced current, we use the right-hand rule. Point your thumb in the direction of flow, and the rest of your digits will extend out the paper and loop in the presence of a magnetic field. Including out magnetization creates a centrifugal movement in the loop, thus according Lenz's law.

The Biot-Savart law allows us to compute the size and direction of a magnetic field produced by a current-carrying wire. at any location where the magnetic field generated by the current-carrying wire segment $\mathrm{螖}\stackrel{鈫�}{s}$is provided

$B=\frac{{渭}_o}{2蟺}\frac{I}{x}$

We utilise Ohm's law to get the induced current I across the loop, as illustrated in the next equation.

$I_{\text{induced}}=\frac{蔚}{R_{\text{loop}}}$

The induced emf, according to Faraday's law, is the change in magnetic flux inside the loop, and it is given by

$蔚=\left|\frac{d{\mathrm{桅}}_{\mathrm{m}}}{dt}\right|$

Where${蠒}_m$ flux loop which the magnetic field passes into area loop$A$

role="math" localid="1648906955738" $\begin{array}{l}d{\mathrm{桅}}_{\mathrm{m}}=dBA=BdA=\left(\frac{{渭}_o}{2蟺}\frac{I}{x}\right)\left(ldx\right)\\ =\frac{{渭}_{o}l}{2蟺}\frac{I}{x}dx.\end{array}$

In station $x=0.5cm$to point $x=0.5\mathrm{cm}+l=0.5\mathrm{cm}+1\mathrm{cm}=1.5\mathrm{cm}$, the permeability changes Where $l$signifies the portal's span. As a response, we incorporate the outflow from $0.5cm$to $1.5cm$.

$鈭�d{\mathrm{桅}}_{\mathrm{m}}=\frac{{渭}_{o}lI}{2蟺}{鈭�}_{0.5}^{1.5}鈥�\frac{1}{x}dx$

${\mathrm{桅}}_{\mathrm{m}}=\frac{{渭}_{O}lI}{2蟺}[\ln 鈦�x{]}_{0.5}^{1.5}$

${\mathrm{桅}}_{\mathrm{m}}=\frac{{渭}_{o}lI}{2蟺}\ln 鈦�3$

Putting ${\mathrm{桅}}_{\mathrm{m}}$phrase into problem to get localid="1648907125510" $蔚$.

localid="1648907395552" role="math" $\begin{array}{l}蔚=\frac{d}{dt}\left(\frac{1.1{渭}_{o}lI}{2蟺}\right)\\ 蔚=\frac{{渭}_{o}l\ln 鈦�3}{2蟺}\frac{dI}{dt}\end{array}$

Since this current $I=I_o{e}^{鈭�t/RC}$, we transfer in this interpretation of localid="1648907162989" $蔚$and localid="1648907169656" $l$into expression to get $I_{\text{induced}}$

localid="1648907385565" $\begin{array}{l}{I}_{\text{induced}}=\frac{{渭}_{o}l\ln 鈦�3}{2蟺R_{\text{loop}}}\frac{d\left(I_o{e}^{鈭�t/RC}\right)}{dt}\\ I_{\text{induced}}=\left(\frac{{渭}_{o}l\ln 鈦�3}{2蟺R_{\text{loop}}}\right)\left(\frac{鈭�I_o}{RC}\right)e^{鈭�t/RC}\end{array}$

Initial value current in circuit,by Ohm's law.

localid="1648907418961" $$\begin{gathered} I_o=\frac{\mathrm{螖}{V}_C}{R} \\ {I}_o=\frac{20\mathrm{V}}{2\mathrm{惟}} \\ {I}_o=10\mathrm{A}. \end{gathered}$$

Feeding the values of $l,{渭}_o,R,\text{loop},C$and $I_o$into expression to get

$I_{\text{induced}}=\left(\frac{{渭}_{o}l\ln 鈦�3}{2蟺R_{\text{loop}}}\right)\left(\frac{鈭�I_o}{RC}\right)e^{鈭�t/RC}$

$=\left(\frac{\left(4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}\right)\left(0.02\mathrm{m}\right)\ln 鈦�3}{2蟺\left(2\mathrm{惟}\right)}\right)\left(\frac{鈭�\left(10\mathrm{A}\right)}{\left(2\mathrm{惟}\right)\left(5脳{10}^{鈭�6}\mathrm{F}\right)}\right)e^{鈭�5渭\mathrm{s}/\left[\right(0.05\mathrm{惟}\left)\right(5渭\mathrm{F}\left)\right]}$

$=53脳{10}^{鈭�3}\mathrm{A}$

localid="1648907456299" $=53mA.$

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