91影视

1. Length$a=2.2cm=0.022m$
2. Width$b=0.80cm=0.080m$
3. Resistance $R=0.40m惟$
4. Current i = 4.7 A
5. Speed v = 3.2 mm/s
6. Distance r = 1.5 b

Use the magnetic flux formula to find the magnitude of magnetic flux. Substituting this value in Faraday鈥檚 law, find the emf induced in the coil. Substituting the value of emf and resistance in Ohm鈥檚 law, find the current induced in the loop.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follows:

$蠒=鈭�\mathrm{Bd}A$

$i_{loop}=\left|\frac{蔚}{R}\right|$

$蔚=-\frac{d蠒}{dt}$

Where,$蠒$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance,饾渶 is emf.

Magnitude of magnetic flux through the loop:

Here, the magnetic flux is generated only due to the long straight wire.

The magnetic field due to the long straight wire is given as,

$B=\frac{{渭}_{0}i}{2蟿蟿r}$

The magnitude of magnetic flux is given as,

$蠒=鈭�\mathrm{Bd}A$

Consider a strip of height dr and the length a, then the area of the strip is,

dA = adr

The distance r varies from $r-\frac{b}{2}$to$r+\frac{b}{2}$

Thus, the magnitude of magnetic flux is,

$\left|蠒\right|=\frac{{渭}_{0}ia}{2蟿蟿}{鈭�}_{r-\frac{b}{2}}^{r+\frac{b}{2}}\frac{1}{r}dr$

$\left|蠒\right|=\frac{{渭}_{0}ia}{2蟿蟿}\ln \left(\frac{r+{\displaystyle \frac{b}{2}}}{r-\frac{b}{2}}\right)$

Since, r = 1.5b , therefore,

$r+\frac{b}{2}=1.5\left(0.80cm\right)+0.4cm=1.6cmandr-\frac{b}{2}=1.5(0.80cm)-0.4cm=0.8cm$

Thus,

$\frac{r+\frac{b}{2}}{r-\frac{b}{2}}=\frac{1.6\mathrm{cm}}{0.8\mathrm{cm}}=2$

Substituting the values,

$\left|\mathrm{蠒}\right|=\frac{\left(4\mathrm{蟿蟿}脳{10}^{-7}\right)\left(4.7\mathrm{A}\right)\left(0.022\mathrm{m}\right)}{2\mathrm{蟿蟿}}\ln \left(2\right)\left|\mathrm{蠒}\right|=1.4脳{10}^{-8}\mathrm{Wb}$

Hence, magnitude of the magnetic flux through the loop is$1.4脳{10}^{-8}Wb$

Current induced in loop :

The induced current in the loop is given by Ohm鈥檚 law,

$i_{loop}=\left|\frac{蔚}{R}\right|$

Where, R is the resistance of the loop and

$蔚=-\frac{d蠒}{dt}$

$$\begin{gathered} 蔚=\frac{{渭}_{0}ia}{2蟿蟿}\left|\frac{d}{dt}\ln \left(\frac{r+{\displaystyle \frac{b}{2}}}{r-{\displaystyle \frac{b}{2}}}\right)\right| \\ 蔚=\frac{{渭}_{0}ia}{2蟿蟿}\left[\frac{1}{r+\frac{b}{2}}-\frac{1}{r-\frac{b}{2}}\right]\frac{dr}{dt} \end{gathered}$$

Since,$\frac{dr}{dt}=v$

Therefore,

$$\begin{gathered} 蔚=\frac{{渭}_{0}ia}{2蟿蟿}\left[-\frac{b}{r^2-{\left({\displaystyle \frac{b}{2}}\right)}^2}\right]v \\ 蔚=\frac{{渭}_{0}iabv}{2蟿蟿\left[r^2-{\left(\frac{b}{2}\right)}^2\right]} \\ {i}_{loop}=\left|\frac{蔚}{R}\right| \\ {i}_{loop}=\frac{{渭}_{0}iabv}{2蟿蟿R\left[r^2-{\left(\frac{b}{2}\right)}^2\right]} \\ {i}_{loop}=\frac{\left(4蟿蟿脳{10}^{-7}{\displaystyle \frac{T.m}{A}}\right)\left(4.7A\right)\left(0.022m\right)\left(0.0080m\right)\left(3.2脳{10}^{-3}{\displaystyle \frac{m}{s}}\right)}{2蟿蟿\left(4脳{10}^{-4}惟\right)\left[2{\left(0.0080m\right)}^2\right]} \\ {i}_{loop}=1脳{10}^{-5}A \end{gathered}$$

Hence, the current induced in loop is$1脳{10}^{-5}A$

Therefore, calculate the magnitude of the magnetic flux using the formula for magnetic flux. Find the current induced in the coil by using Faraday鈥檚 law and Ohm鈥檚 law.