Question

The relation between the charge flow \(\Delta Q\) through the circuit of resistance \(r\) and the change in the magnetic flux \(\Delta \phi_B\) is

• A. \(\Delta Q=\dfrac{\Delta \phi_B}{r}\)
• B. \(\Delta \phi_B=\dfrac{\Delta Q}{r}\)
• C. \(\Delta \phi_B=\Delta Q\)
• D. \(\Delta \phi_B=-\dfrac{r\Delta Q}{q}\)
• E. \(\Delta Q=-\dfrac{\Delta \phi_B}{r}\)

Hint:

For total induced charge: \[ Q=\frac{\Delta \phi}{R} \] This is a standard one-step result from Faraday's law and Ohm's law.

Answer 1

The Correct Option is A

Induced emf: \[ \mathcal{E}=-\frac{d\phi_B}{dt} \] Current: \[ I=\frac{\mathcal{E}}{r} \] So, \[ I=\frac{1}{r}\left(-\frac{d\phi_B}{dt}\right) \] Since \[ I=\frac{dQ}{dt} \] \[ \frac{dQ}{dt}=-\frac{1}{r}\frac{d\phi_B}{dt} \] Ignoring sign for magnitude relation: \[ \Delta Q=\frac{\Delta \phi_B}{r} \] Hence, \[ \boxed{(A)\ \Delta Q=\frac{\Delta \phi_B}{r}} \]