Question

A rectangular metal loop with $0.050$ resistance is placed next to one wire of the RC circuit shown in $FIGUREP30.52$. The capacitor is charged to $2V$with the polarity shown, then the switch is closed at $t=0s$.

a. What is the direction of current in the loop for $t>0s$?

b. What is the current in the loop at $t=5.0\mu s$? Assume that

only the circuit wire next to the loop is close enough to produce a significant magnetic field.

Short answer

Part $\left(a\right)$

$\left(a\right)$The current with in loop is flowing clockwise.

Part $\left(b\right)$

$\left(b\right)$The current in loop is $I_{\text{induced}}=53\mathrm{mA}.$

Step-by-step solution

Step: 1  Direction of current in loop: (part a)

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.

Step: 2  Magnitude and direction:  (part b)

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 $\mathrm{\Delta }\overrightarrow{s}$is provided

$B=\frac{{\mu }_o}{2\pi }\frac{I}{x}$

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

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

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

$\epsilon =\left|\frac{d{\mathrm{\Phi }}_{\mathrm{m}}}{dt}\right|$

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

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

Step: 3 Substituting values: (part b)

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 $1.5cm$.

$\int d{\mathrm{\Phi }}_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }{\int }_{0.5}^{1.5} \frac{1}{x}dx$

${\mathrm{\Phi }}_{\mathrm{m}}=\frac{{\mu }_{O}lI}{2\pi }[\ln x{]}_{0.5}^{1.5}$

${\mathrm{\Phi }}_{\mathrm{m}}=\frac{{\mu }_{o}lI}{2\pi }\ln 3$

Step : 4 Finding loop current: (part b)

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

localid="1648907395552" role="math" $\begin{array}{l}\epsilon =\frac{d}{dt}\left(\frac{1.1{\mu }_{o}lI}{2\pi }\right)\\ \epsilon =\frac{{\mu }_{o}l\ln 3}{2\pi }\frac{dI}{dt}\end{array}$

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

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

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

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

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

$I_{\text{induced}}=\left(\frac{{\mu }_{o}l\ln 3}{2\pi R_{\text{loop}}}\right)\left(\frac{-I_o}{RC}\right)e^{-t/RC}$

$=\left(\frac{\left(4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}\right)\left(0.02\mathrm{m}\right)\ln 3}{2\pi \left(2\mathrm{\Omega }\right)}\right)\left(\frac{-\left(10\mathrm{A}\right)}{\left(2\mathrm{\Omega }\right)\left(5\times {10}^{-6}\mathrm{F}\right)}\right)e^{-5\mu \mathrm{s}/\left[\right(0.05\mathrm{\Omega }\left)\right(5\mu \mathrm{F}\left)\right]}$

$=53\times {10}^{-3}\mathrm{A}$

localid="1648907456299" $=53mA.$

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