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.
i don't understand physics
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.
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
To find the magnitude of the induced potential difference in the solenoid, let's break down the steps.
Step 1: Calculate the induced electromotive force (emf) using Faraday's Law of Electromagnetic Induction:
- Faraday's Law states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop.
- In this case, the closed loop is the solenoid with N turns, and the magnetic field B is changing at a rate of dB/dt.
- The magnetic flux through the solenoid is given by Φ = B*A, where A is the cross-sectional area of the solenoid.
- The induced emf is therefore given by emf = -N*dΦ/dt = -N*d(B*A)/dt.
- Since the magnetic field B is confined within the solenoid and the cross-sectional area A is constant, we can simplify this equation to emf = -N*A*dB/dt.
Step 2: Calculate the induced potential difference (Vind) using Ohm's Law:
- The induced potential difference is given by Vind = -emf = N*A*dB/dt.
Now, let's plug in the given values:
- N = 200 turns.
- A = 60 cm^2 = 60 * 10^(-4) m^2 (converted to square meters).
- dB/dt = 0.20 T/s.
Substituting these values into the equation for Vind, we get:
Vind = N*A*dB/dt
Vind = (200)*(60 * 10^(-4))*(0.20)
Vind = 0.24 V
Therefore, the magnitude of the induced potential difference in the solenoid is 0.24V (option d).