How Does a Flywheel Power a 40 T Magnetic Field?

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A research lab is attempting to generate a 40 T magnetic field using energy from a flywheel, which consists of circular plates rotating at 500 RPM. The energy stored in the magnetic field is calculated to be 8 MJ, while the initial energy of the flywheel is approximately 1.04 x 10^9 J. To determine the reduction in angular velocity required to supply the necessary energy, the conservation of energy principle is applied. The calculation shows that the flywheel's angular velocity must decrease by 20.64 RPM to achieve the energy conversion needed for the magnetic field. This highlights the relationship between rotational energy and electromagnetic energy in the context of magnetic field generation.
Nylex
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Can anyone help me with this?

"A research lab has a magnetic facility, which can produce intense, pulsed magnetic fields. In 1 experiment, a B-field of 40 T is to be produced over a cylindrical volume, 40 cm in length and 20 cm in diameter.

The energy required to pulse the magnet is supplied by a flywheel, constructed from a set of circular plates, 5 m in diameter and with a total mass of 10^5 kg, rotating at 500 RPM."

Calculate how much energy is stored in the B-field

For this, I used magnetic energy density, u = U/V = (B^2)/2μ0, where U = energy stored, V = volume in field.

U = 8 MJ

Estimate the reduction in angular velocity of the flywheel needed in order to supply energy necessary to produce the required B-field. Assume conversion from rotational energy to magnetic energy is 100% efficient.

Here I'm stuck. If the conversion is 100% efficient, why can't I use KE(rot) = U, with KE(rot) = (1/2)Iω^2?
 
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Nylex said:
Here I'm stuck. If the conversion is 100% efficient, why can't I use KE(rot) = U, with KE(rot) = (1/2)Iω^2?

It's asking for the change in the angular speed - the wheel doesn't necessarily stop:

\frac{1}{2} I \omega_0^2= \frac{1}{2} I \omega_f^2 + E_{magnet}
 


To answer your question, you cannot simply equate the kinetic energy of the flywheel to the energy stored in the magnetic field because the two systems have different forms of energy. The kinetic energy of the flywheel is in the form of rotational energy, while the energy stored in the magnetic field is in the form of electromagnetic energy. Therefore, you cannot directly compare the two energies.

To estimate the reduction in angular velocity of the flywheel, you will need to consider the conservation of energy. The initial energy of the flywheel, which is in the form of rotational energy, will be converted into electromagnetic energy to produce the magnetic field. This conversion will result in a decrease in the flywheel's rotational energy and therefore a decrease in its angular velocity.

To calculate the reduction in angular velocity, you can use the equation:

Δω = (ΔE)/(Iω)

Where ΔE is the change in energy, I is the moment of inertia of the flywheel, and ω is the initial angular velocity.

Using the given values, you can calculate the initial energy of the flywheel:

E = (1/2)Iω^2 = (1/2)(10^5 kg)(500 RPM)^2 = 1.04 x 10^9 J

Then, using the energy stored in the magnetic field calculated earlier (8 MJ = 8 x 10^6 J), you can calculate the change in energy:

ΔE = 1.04 x 10^9 J - 8 x 10^6 J = 1.032 x 10^9 J

Finally, plug in the values into the equation to calculate the reduction in angular velocity:

Δω = (1.032 x 10^9 J)/[(10^5 kg)(500 RPM)] = 20.64 RPM

Therefore, the flywheel's angular velocity will need to decrease by 20.64 RPM in order to supply the necessary energy to produce the required magnetic field.
 
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