- #1
Silversonic
- 121
- 1
It is for my university course but I didn't find it worthy of the advanced physics section.
A circular metallic disk of radius b rotates about an axis perpendicular to
its centre with an angular velocity ω in a constant magnetic field,
strength B, which lies in the same direction as the rotation axis. A potential difference is generated between the centre of the disc and its edge.
The disc is connected to an external circuit. Assume that the external circuit has resistance R and that a current I flows through it (and therefore through the disk). Calculate the rate of change of the disk’s kinetic energy.
It gives me a hint to first calculate the electrical power dissipated in the resistor and the torque acting on the disk.
Power dissipated = R*I^2
Rotational Kinetic energy = 0.5*I*ω^2
Torque = Ia = rF (a is the angular acceleration, equal to dw/dt).
Force (due to magnetic field) on the current in the disc = BIb.
E.M.F produced within the disc = 0.5*ω*B*a^2
Also. As is known, a current is induced in the Faraday disc and as this current is in a radial direction it experiences an extra force due to the magnetic field. This extra force is in such a direction as to oppose the angular motion. So how would I calculate the overall torque of the system due to this extra motion of the charged particles? I don't think it could simply be T = rF, would it not be the integral of rF with limis b and 0?
Next thing, the change in the rotational kinetic energy with time is going to be dE/dt.
This means
dE/dt = 0.5*I*a^2, as dw/dt = a, the angular acceleration.
But the only thing is F = BIb, surely this will be changing with time too. If the angular velocity of the disc decreases the amount of flux lines it cuts per second also decreases. This decreases the e.m.f produced and means there is a reduced current. So the current is also dependent on time.Ugh, there are so many factors to consider. Can anyone tell me where to start?
Homework Statement
A circular metallic disk of radius b rotates about an axis perpendicular to
its centre with an angular velocity ω in a constant magnetic field,
strength B, which lies in the same direction as the rotation axis. A potential difference is generated between the centre of the disc and its edge.
The disc is connected to an external circuit. Assume that the external circuit has resistance R and that a current I flows through it (and therefore through the disk). Calculate the rate of change of the disk’s kinetic energy.
Homework Equations
It gives me a hint to first calculate the electrical power dissipated in the resistor and the torque acting on the disk.
Power dissipated = R*I^2
Rotational Kinetic energy = 0.5*I*ω^2
Torque = Ia = rF (a is the angular acceleration, equal to dw/dt).
Force (due to magnetic field) on the current in the disc = BIb.
E.M.F produced within the disc = 0.5*ω*B*a^2
The Attempt at a Solution
I'm very confused with many aspects to this question. Firstly how is the electrical power dissipated in the resistor even linked to the rotational kinetic energy of the disc?Also. As is known, a current is induced in the Faraday disc and as this current is in a radial direction it experiences an extra force due to the magnetic field. This extra force is in such a direction as to oppose the angular motion. So how would I calculate the overall torque of the system due to this extra motion of the charged particles? I don't think it could simply be T = rF, would it not be the integral of rF with limis b and 0?
Next thing, the change in the rotational kinetic energy with time is going to be dE/dt.
This means
dE/dt = 0.5*I*a^2, as dw/dt = a, the angular acceleration.
But the only thing is F = BIb, surely this will be changing with time too. If the angular velocity of the disc decreases the amount of flux lines it cuts per second also decreases. This decreases the e.m.f produced and means there is a reduced current. So the current is also dependent on time.Ugh, there are so many factors to consider. Can anyone tell me where to start?