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.
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.
Answers
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
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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