Question

The magnetic flux passing through a metal ring varies with time t as \[{{\phi }_{B}}=3(a{{t}^{3}}-b{{t}^{2}})T.{{m}^{2}}\] with \[a=2.00{{s}^{-3}}\text{ and b}=6.00{{s}^{-2}}\]. The resistance of the ring is \[3.0\Omega \]. Determine the maximum current induced in the ring during the interval from \[t=0s\text{ to t}=2.0s\].

Answer 1

Hint: We are given the magnetic flux equation due to the metal ring as a function of time. We can find the induced electromotive force (emf) due to this magnetic flux. From that we can find the induced current using appropriate relations.

Complete answer:
We need to refresh our memory on the relations between a magnetic time-varying flux and its impact on a conductor. We know that such a time-dependent flux will induce an electric field in the conductor due to an induced electromotive force (emf).
According to Faraday’s law of Electromagnetic induction and Lenz’s law, the magnitude of induced emf \[\varepsilon \] in a circuit is equal to the rate of change of magnetic flux through the circuit.
i.e.,
\[\left| \varepsilon \right|=-N\left| \dfrac{d{{\phi }_{B}}}{dt} \right|\]
where N is the number of turns in a coil.
In the given situation, N is 1. The rate of change of magnetic flux is given as \[{{\phi }_{B}}=3(a{{t}^{3}}-b{{t}^{2}})T.{{m}^{2}}\]. So, let us find the induced emf in the ring –
 i.e.,
\[\begin{align}
  & \left| \varepsilon \right|=\dfrac{d}{dt}\left( 3(a{{t}^{3}}-b{{t}^{2}}) \right) \\
 & \left| \varepsilon \right|=9a{{t}^{2}}-6bt \\
 & \text{given},\text{ a}=2{{s}^{-3}}\text{ and b}=6{{s}^{-2}} \\
 & \text{Substituting,} \\
 & \varepsilon =(9\times 2\times {{t}^{2}})-(6\times 6\times t) \\
 & \varepsilon =18{{t}^{2}}-36t \\
\end{align}\]
Now, let us find the induced current at time ‘t’ as –
\[\begin{align}
  & i=\dfrac{\varepsilon }{r} \\
 & \text{Given, r}=3\Omega \\
 & \Rightarrow i=\dfrac{18{{t}^{2}}-36t}{3} \\
 & \Rightarrow i=6{{t}^{2}}-12t \\
\end{align}\] --(1)
We can find the value of ‘t’ by maximizing the induced current. i.e., the current will be maximum at \[\dfrac{di}{dt}=0\]
Applying this to (1),
\[\begin{align}
  & \dfrac{di}{dt}=0 \\
 & \Rightarrow \dfrac{di}{dt}=\dfrac{d}{dt}(6{{t}^{2}}-12t)=0 \\
 & \Rightarrow 12t-12=0 \\
 & \Rightarrow \text{ }t=\dfrac{12}{12}=1s \\
\end{align}\]
So, at time \[t=1\], the maximum induced current will be observed.
Therefore, we can find the maximum induced current by substituting \[t=1\]in (1),
i.e.,
\[\begin{align}
  & i=6{{t}^{2}}-12t \\
 & \Rightarrow \text{ i}=(6\times {{\left( 1 \right)}^{2}})-(12\times 1) \\
 & \Rightarrow \,\text{ }i=6-12 \\
 & \Rightarrow \text{ }i=6A \\
\end{align}\]

The maximum induced current in the circuit is 6A.

Note:
We can find the maximum induced current by this method or without substituting the values of a and b initially. We can keep it as variable till the last equation of induced current and substitute it there according to your ease of computing.