The solution to the questions are given as
- [tex]t=40.39 \mathrm{sec}[/tex]
- [tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]
- the direction of induced current will be Counterclock vise.
What is the direction of the current induced in the loop, as viewed from above the loop.?
Given, $B(t)=(1.4 T) e^{-0.057 t}$
[tex]$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}[/tex]
[tex]\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$[/tex]
[tex]\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}[/tex]
[tex]\varepsilon &=(0.12v)e^{0.057t}[/tex]
(b) [tex]Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$[/tex]
[tex]\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}[/tex]
c)
In conclusion, the direction of the induced current will be Counterclockwise.
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