Question

Flux $\phi $ (in Weber) in a closed circuit of resistance $10\Omega $ varies with time t (in seconds) according to the equation $\phi = 6{t^2} - 5t + 1$ . What the magnitude of the induced current in $0.25$ seconds:
A. $0.8A$
B. $1.2A$
C. $0.6A$
D. $0.2A$

Answer 1

Hint:In this question, we had used the basic formulas. Firstly, we had calculated the emf which is differential of the flux (which is given in the question) then taken the emf same as terminal potential difference because no current flows then find the induced current by ohm’s law.

Complete step by step answer:
We are given that the flux in a circuit is $\phi = 6{t^2} - 5t + 1$. We have to find the current induced through the circuit. And the additional values provided are the resistance and time period. So, firstly finding the electromotive force or emf, the relation used is that the emf is the derivative of the flux with respect to time.
$emf = \dfrac{{d\phi }}{{dt}}$
Finding emf by using the given values,
$
emf = \dfrac{{d\left( {6{t^2} - 5t + 1} \right)}}{{dt}} \\
\Rightarrow emf = 12t - 5 \\
$
Now, using the given time $t = 0.25\sec $
$
emf = 12\left( {0.25} \right) - 5 \\
\Rightarrow emf = - 2 \\
\Rightarrow \left| {emf} \right| = \left| { - 2} \right| = 2 \\ $
Emf is basically the potential difference so, in this case we can take the use of ohm’s law
$
I = \dfrac{{emf}}{R} \\
\therefore I = \dfrac{2}{{10}} = 0.2A \\ $
So, the current induced is $0.2A$ and the correct option is D.

Note:Electromotive force is taken as the potential difference when no current is flowing through it. But these two quantities are not the same. Basic definition of emf is that it is the energy provided by the battery for each charge flowing through it.