Question

when the north pole of a bar magnet is moved into a solenoid, the needle on a galvanometer attached to the solenoid moves to the left, indicating that a current is produced. the magnet is then pulled out at a faster speed. what will be the result? a larger current in the same direction a larger current in the opposite direction a smaller current in the same direction a smaller current in the opposite direction

Explanation:

Step1: Recall Faraday's law

According to Faraday's law of electromagnetic - induction, the induced emf $\epsilon=-N\frac{\Delta\Phi}{\Delta t}$, where $N$ is the number of turns in the solenoid, $\Delta\Phi$ is the change in magnetic flux, and $\Delta t$ is the time interval.

Step2: Analyze the pulling - out situation

When the magnet is pulled out, the magnetic flux through the solenoid changes in the opposite direction compared to when it is pushed in. So, the direction of the induced current will be opposite.

Step3: Consider the speed effect

Since the magnet is pulled out at a faster speed, $\Delta t$ is smaller. A smaller $\Delta t$ for the same $\Delta\Phi$ (assuming the same magnet - solenoid system) results in a larger induced emf according to $\epsilon=-N\frac{\Delta\Phi}{\Delta t}$. A larger induced emf means a larger induced current.

Answer:

a larger current in the opposite direction