Asked by Jasmine

A solenoid with 200 turns and a cross sectional area of 60 cm^2 has a magnetic field of .60 T along its axis. If the field is confined within the solenoid and changes at a rate of .20 T/s, the magnitude of the induced potential difference in the solenoid will be:
a) .00020V
b).02V
c).001V
d).24V

Ok, so I know that when a magnetic field is changing and the area is constant the equation is:

Vind -Acos(theta)dB/dt

I thought that since we had the rate of of change in B and the area those were the only things I needed but I get 1.2E-3 and the answer is supposed to be .24V but I don't know how to get that. And I don't know how to relate the number of turns to the equation.
Answered by Jasmine
So now I noticed that, if I multiply the number of turns to my previous answer, I get .24 V. But I don't know why that must be done.
Answered by Damon
each and every turn gets a voltage of .0012
the turns are end to end, in series, so you add voltages, multiply by 200 to get .24
Answered by mahra
i don't understand physics
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