Question

The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \Phi $ in a time $\Delta t$ . Then the total quantity of electric charge $Q$ that passes any point in the circuit during the time $\Delta t$ is represented by:
A. $Q = \dfrac{{\Delta \phi }}{{\Delta t}}$
B. $Q = \dfrac{{\Delta \phi }}{R}$
C. $Q = R.\dfrac{{\Delta \phi }}{{\Delta t}}$
D. $Q = \dfrac{1}{R}\dfrac{{\Delta \phi }}{{\Delta t}}$

Answer 1

Hint: To solve this type of question, one must know about the concept of Faraday’s law of EMI. By using this concept, we will find the total amount of charges passing through the circuit by substituting in the formula and solving it to get the required solution.

Formula used:
$e = \dfrac{{\Delta \phi }}{{\Delta t}}$
Where,
$e$ is the induced voltage,
$\Delta \Phi $ is the change in magnetic flux and
$\Delta t$ is the change in time.

Complete answer:
From Faraday’s law of EMI,emf induced in the circuit is given by,
$e = \dfrac{{\Delta \phi }}{{\Delta t}}$
And if $R$ is the resistance in the circuit then it becomes,
$I = \dfrac{e}{R}$
$ \Rightarrow I = \dfrac{{\Delta \phi }}{{\Delta t.R}}$
So, the total amount of charge passing through the circuit will become,
$
  \because Q = I \times \Delta t \\
   \Rightarrow Q = \dfrac{{\Delta \phi }}{{\Delta t.R}}.\Delta t \\
   \Rightarrow Q = \dfrac{{\Delta \phi }}{R} \\
 $
So, the total amount of charge passing through the circuit is given by $\dfrac{{\Delta \phi }}{R}$ .
Hence, the correct option is B.

Note:
Whenever the magnetic flux linked with a circuit changes an emf is induced in the circuit. The emf stays in the circuit as long as the flux keeps changing. The nature of the induced emf is such that it opposes the cause due to how it is produced, i.e., the flux changes.