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pc2-brazil
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This is for self-study.
The armature of a motor has 97 turns each of area 190 cm² and rotates in a uniform magnetic field of 0.33 T. A potential difference of 24 V is applied. If no load is attached and friction is neglected, find the rotational speed at equilibrium.
Initially, the potential difference is 24 V. Because of this potential difference, current starts flowing through the motor. Then, because of the 0.33 T magnetic field, the motor starts rotating. The rotational motion changes the magnetic flux through the motor's coil, which produces an induced EMF that opposes the original EMF. If we call N the number of turns, A the area of the motor's coil, and θ the angle from the magnetic field vector B to the vector normal to the plane of the coil, then the magnetic flux through the coil is:
[tex]\Phi=NAB\cos{\theta}[/tex]
So, the induced EMF is:
[tex]\varepsilon_{ind}=-\frac{\mathrm{d} }{\mathrm{d} t}(NAB\cos{\theta})=NAB\frac{\mathrm{d} \theta }{\mathrm{d} t}\sin{\theta}[/tex]
If we call ω the angular velocity as a function of time, and consider that the angle in t = 0 is 0º, we have that θ = ωt, so the net EMF in function of time is:
[tex]\varepsilon = 24- NAB\omega\sin\omega t[/tex]
Is this correct so far? I'm not sure how to proceed from here, but I think that the rotational speed in equilibrium is reached when the angular acceleration α = 0.
Thank you in advance.
Homework Statement
The armature of a motor has 97 turns each of area 190 cm² and rotates in a uniform magnetic field of 0.33 T. A potential difference of 24 V is applied. If no load is attached and friction is neglected, find the rotational speed at equilibrium.
Homework Equations
The Attempt at a Solution
Initially, the potential difference is 24 V. Because of this potential difference, current starts flowing through the motor. Then, because of the 0.33 T magnetic field, the motor starts rotating. The rotational motion changes the magnetic flux through the motor's coil, which produces an induced EMF that opposes the original EMF. If we call N the number of turns, A the area of the motor's coil, and θ the angle from the magnetic field vector B to the vector normal to the plane of the coil, then the magnetic flux through the coil is:
[tex]\Phi=NAB\cos{\theta}[/tex]
So, the induced EMF is:
[tex]\varepsilon_{ind}=-\frac{\mathrm{d} }{\mathrm{d} t}(NAB\cos{\theta})=NAB\frac{\mathrm{d} \theta }{\mathrm{d} t}\sin{\theta}[/tex]
If we call ω the angular velocity as a function of time, and consider that the angle in t = 0 is 0º, we have that θ = ωt, so the net EMF in function of time is:
[tex]\varepsilon = 24- NAB\omega\sin\omega t[/tex]
Is this correct so far? I'm not sure how to proceed from here, but I think that the rotational speed in equilibrium is reached when the angular acceleration α = 0.
Thank you in advance.