A 695-loop square armature coil with a side of 15.0 cm rotatesat 55.0 rev/s in a

Question

A 695-loop square armature coil with a side of 15.0 cm rotatesat 55.0 rev/s in a uniform magnetic field of strength 0.7 T. Whatis the rms voltage output of the generator?

The equation I used was: V = N*B*A*omega*sin(omega*t).

sin(omega*t) comes out to 1 in order to get peak voltage, so we cancancel it out, which left V_peak = N*B*A*omega.

So now I went about converting everything and getting it ready toplug into the equation.

For N, I have 695 loops. For B, the magnetic field is given at .7T. For A, the area, I have pi*(.15m)^2. And for omega, I have 55rev/s * (2*pi) which comes out to 345.6 rad/s. Plugging all thisinto the equation gave me 1.188E4 V.

Now we have to get it into V_rms which is V_peak * 1/sqrt(2).Solving for this I got 8.403E3 V. This answer was incorrect. Any help would very much be appreciated! A 695-loop square armature coil with a side of 15.0 cm rotatesat 55.0 rev/s in a uniform magnetic field of strength 0.7 T. Whatis the rms voltage output of the generator?

The equation I used was: V = N*B*A*omega*sin(omega*t).

sin(omega*t) comes out to 1 in order to get peak voltage, so we cancancel it out, which left V_peak = N*B*A*omega.

So now I went about converting everything and getting it ready toplug into the equation.

For N, I have 695 loops. For B, the magnetic field is given at .7T. For A, the area, I have pi*(.15m)^2. And for omega, I have 55rev/s * (2*pi) which comes out to 345.6 rad/s. Plugging all thisinto the equation gave me 1.188E4 V.

Now we have to get it into V_rms which is V_peak * 1/sqrt(2).Solving for this I got 8.403E3 V. This answer was incorrect. Any help would very much be appreciated!

Explanation / Answer

   Only correction required is that coil ofarmature is square, not circular, so the area will be    A   =   0.152          =   0.0225   m2          =   0.0225   m2

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