Consider a magnet surrounded by a wire loop with an on/off switch $S$ as shown in the figure. If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.
(N/A) No, a current will not flow in the circuit.
According to Faraday's law of electromagnetic induction, an induced electromotive force $(EMF)$ and consequently an induced current are produced only when there is a change in the magnetic flux linked with the coil.
In this scenario, the magnet is stationary relative to the wire loop. Even when the switch $S$ is closed, the magnetic field lines passing through the loop remain constant. Since there is no change in the magnetic flux $(\Delta \Phi = 0)$, no induced current is generated in the circuit.
According to Faraday's law of electromagnetic induction, an induced electromotive force $(EMF)$ and consequently an induced current are produced only when there is a change in the magnetic flux linked with the coil.
In this scenario, the magnet is stationary relative to the wire loop. Even when the switch $S$ is closed, the magnetic field lines passing through the loop remain constant. Since there is no change in the magnetic flux $(\Delta \Phi = 0)$, no induced current is generated in the circuit.
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