A magnet is moved towards a coil,an emf and c | vedclass

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Question

(C) According to Faraday's law of electromagnetic induction,the induced $emf$ is given by $\epsilon = -\frac{d\phi}{dt}$.When the magnet is moved slow

Vedclass · Accepted Answer

$emf$ will decrease,but charge will remain unchanged.

Vedclass · Answer

(C) According to Faraday's law of electromagnetic induction,the induced $emf$ is given by $\epsilon = -\frac{d\phi}{dt}$.When the magnet is moved slowly,the rate of change of magnetic flux $\frac{d\phi}{dt}$ decreases,which causes the induced $emf$ to decrease.The induced charge $q$ is given by $q = \int I dt = \int \frac{\epsilon}{R} dt = \int \frac{1}{R} \frac{d\phi}{dt} dt = \frac{\Delta\phi}{R}$.Since the total change in magnetic flux $\Delta\phi$ and the resistance $R$ of the coil remain the same,the induced charge $q$ remains unchanged.

$A$ magnet is moved towards a coil,an $emf$ and charge are induced in it. If the magnet is moved slowly,how will the new $emf$ and charge be affected?

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