Calculating Induced Voltage in a Rotating Rectangular Loop

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A rectangular loop measuring 40cm by 30cm rotates at 130 r/s in a magnetic field of 0.06 T, leading to the calculation of induced voltage. The approach involved using the formula B dot ds, with the sides of the rectangle and the cosine of the angular frequency in the normal direction. The calculation yielded an induced voltage of 46.8 sin(130t), while the textbook answer is -54. The discussion confirms that the induced voltage must oscillate due to the unchanging magnetic field. Verification of the calculations and the nature of the induced voltage is sought.
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A 40cmX30cm rectangular loop rotates at 130 r/s in a magnetic field of .06 in the direction normal to the axis of rotation. If the loop has 50 turns, determine the induced voltage in the loop.

Heres what I did. I used B dot ds. I had my ds being the sides of the rectangle and coswt in the normal direction. I then took the derivative of that. So I multiplied 130*.4*.3*50*.06. The 50 comes from the number of turns.
I ended up with an answer of 46.8 sin 130 t. The book has an answer of -54. Since the magnetic field is unchanging, the voltage has to be time-varying.
 
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can somebody confirm my answer?
 
Looks right, and yes, the voltage should definitely oscillate.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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