Question

The magnetic flux passing through a metal ring varies with time $t$ as ${\varphi }_B=3(a{t}^3-b{t}^2){\text{Tm}}^2$ with $a=2.00{\text{s}}^{-3}$ and $b=6.00{\text{s}}^{-2}$. The resistance of the ring is $3.0\Omega .$ The maximum current ( $\text{in A}$) induced in the ring during the time interval $t=0$ to $t=2.0\text{s}$ is

Answer 1

Given, ${\varphi }_B=3(a{t}^3-b{t}^2)$

According to Faraday’s law of electromagnetic induction, the magnitude of induced $\text{emf}$ in a circuit is,

$\left|\begin{array}{c}E\end{array}\right|=\left|\begin{array}{c}\frac{d{\varphi }_B}{dt}\end{array}\right|=3(3a{t}^2-2bt)$

Hence, induced current be,

$i=\frac{\left|\begin{array}{c}E\end{array}\right|}{R}=\frac{3(3a{t}^2-2bt)}{3}=3a{t}^2-2bt$

Putting the value of $a$ and $b$ we get,

$i=6t^2-12t$

For current to be maximum,

$\frac{di}{dt}=0$

$\frac{d}{dt}(6t^2-12t)=0\Rightarrow 12t=12$

$\therefore t=1\text{s}$

i.e. at $t=1\text{s}$, current is maximum. So, the magnitude of the maximum current is,

$\left|\begin{array}{c}{i}_{max}\end{array}\right|=|6(1{)}^2-12(1)||=6\text{A}$

Accepted answers: $6\text{or}6.0\text{or}6.00$