Calculating Induced Potential Difference in a Bar Magnet and Coil Setup

[XxphysicsxX]

Homework Statement

A baar magnet inserted into a coil of 500 turns induces a potential difference of 1.5 V across the coil. Determine the induced potential difference when;

a> 250 turn coilis used.
b> the bar magnet is moved twice as fast through the coil.
c> three identical magnets are inserted at once, side by side
d> all three of the above charges occur simultaneously.

The Attempt at a Solution

a> 1.5V/ 500 = V/ 250
solve for V, V= 0.75 V

b> I am not quite sure, I am guessing there will be no difference in the voltage?

c> (0.75V)(3) = 2.25V

d> do I just add the voltages from a, b, and c?

I would greatly appreciate it if anyone could confirm the answers I got or let me know if I am on the right track with b and d?

Thankyou in advance!

 

[anathema]

I can confirm that your answers for a, c, and d are correct. For b, you are correct that there will be no difference in the induced potential difference. This is because the induced voltage is directly proportional to the number of turns in the coil, and the speed at which the magnet moves does not affect the number of turns in the coil. Therefore, the induced voltage will remain at 1.5 V regardless of the speed of the magnet.

For d, you are correct that you can simply add the individual voltages from a, b, and c to get the total induced potential difference. This is because each of these factors (number of turns, speed of magnet, and number of magnets) affect the induced voltage independently, so you can simply add them together to get the total effect.

Overall, you have a good understanding of the concept of induced potential difference and how it is affected by different factors. Keep up the good work!