MHT CET · Physics · Electromagnetic Induction
The magnetic flux through a coil of resistance \(R\) changes by an amount \(\Delta \phi\) in time \(\Delta t\). The total quantity of induced electric charge \(Q\) is
- A \(\frac{\Delta \phi}{\Delta t}\)
- B \(-\frac{\Delta \phi}{\Delta t}+R\)
- C \(\frac{\Delta \phi}{R}\)
- D \(\frac{\Delta \phi}{\Delta t} \times R\)
Answer & Solution
Correct Answer
(D) \(\frac{\Delta \phi}{\Delta t} \times R\)
Step-by-step Solution
Detailed explanation
From Faraday's law of EMI, emf induced in the circuit is given by,
\(e=-\frac{\Delta \phi}{\Delta t}\)
And if \(R\) is the resistance in the circuit then it becomes,
\(\begin{aligned} & I=\frac{e}{R} \\ & \quad \Rightarrow I=-\frac{\Delta \phi}{\Delta t \cdot R}\end{aligned}\)
So, the total amount of charge passing through the circuit will become,
\(\begin{aligned} & \because \Delta Q=I \times \Delta t \\ & \Rightarrow \Delta Q=-\frac{\Delta \phi}{\Delta t \cdot R} \cdot \Delta t \\ & \Rightarrow \Delta Q=-\frac{\Delta \phi}{R}\end{aligned}\)
So, the total amount of charge passing through the circuit is given by \(\frac{\Delta \phi}{R}\)
\(e=-\frac{\Delta \phi}{\Delta t}\)
And if \(R\) is the resistance in the circuit then it becomes,
\(\begin{aligned} & I=\frac{e}{R} \\ & \quad \Rightarrow I=-\frac{\Delta \phi}{\Delta t \cdot R}\end{aligned}\)
So, the total amount of charge passing through the circuit will become,
\(\begin{aligned} & \because \Delta Q=I \times \Delta t \\ & \Rightarrow \Delta Q=-\frac{\Delta \phi}{\Delta t \cdot R} \cdot \Delta t \\ & \Rightarrow \Delta Q=-\frac{\Delta \phi}{R}\end{aligned}\)
So, the total amount of charge passing through the circuit is given by \(\frac{\Delta \phi}{R}\)
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