Q188 / 200 · MHT CET - Medical 2026 (21 April Shift 2) · Physics
The magnetic flux through a coil of resistance $R$ changes by an amount $\Delta\Phi$ in the time $\Delta t$. The total quantity of electric charge ($Q$) induced is
- A$Q = -\dfrac{\Delta\Phi}{\Delta t} + R$
- B$Q = \dfrac{\Delta\Phi}{\Delta t} \times R$
- Correct answerC$Q = \dfrac{\Delta\Phi}{R}$
- D$Q = \dfrac{\Delta\Phi}{\Delta t}$
Solution
According to Faraday's law of electromagnetic induction, the magnitude of induced emf is given by
$|e| = \dfrac{\Delta\Phi}{\Delta t}$
The induced current in the coil of resistance $R$ is
$I = \dfrac{|e|}{R} = \dfrac{\Delta\Phi}{R \Delta t}$
The total quantity of electric charge $Q$ induced in time $\Delta t$ is
$Q = I \Delta t = \left(\dfrac{\Delta\Phi}{R \Delta t}\right) \Delta t = \dfrac{\Delta\Phi}{R}$
Answer: $Q = \dfrac{\Delta\Phi}{R}$
$|e| = \dfrac{\Delta\Phi}{\Delta t}$
The induced current in the coil of resistance $R$ is
$I = \dfrac{|e|}{R} = \dfrac{\Delta\Phi}{R \Delta t}$
The total quantity of electric charge $Q$ induced in time $\Delta t$ is
$Q = I \Delta t = \left(\dfrac{\Delta\Phi}{R \Delta t}\right) \Delta t = \dfrac{\Delta\Phi}{R}$
Answer: $Q = \dfrac{\Delta\Phi}{R}$
Hence, the correct answer is option: 'C'.