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Day 266 Solution: Ashmit Dutta

The document summarizes the solution to calculating the energy of a rotating charged cylinder. It first calculates the bound charge density and surface charge density of the cylinder. It then uses Gauss's law to find the electric field and calculates the electric potential energy of the stationary cylinder. Next, it determines the volume current density and uses Ampere's law to find the magnetic field generated by the rotating cylinder. It calculates the magnetic potential energy and combines it with the electric potential energy to give the total energy of the rotating cylinder, which is higher due to the work done to make it rotate.

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693 views2 pages

Day 266 Solution: Ashmit Dutta

The document summarizes the solution to calculating the energy of a rotating charged cylinder. It first calculates the bound charge density and surface charge density of the cylinder. It then uses Gauss's law to find the electric field and calculates the electric potential energy of the stationary cylinder. Next, it determines the volume current density and uses Ampere's law to find the magnetic field generated by the rotating cylinder. It calculates the magnetic potential energy and combines it with the electric potential energy to give the total energy of the rotating cylinder, which is higher due to the work done to make it rotate.

Uploaded by

Obama binladen
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Day 266 Solution

Ashmit Dutta
June 2020

Solution
The bound charge density of the cylinder is given as:

1 ∂(r · kr)
ρb = −∇ · P = −∇(kr) = − = −2k.
r ∂r
There must be a surface charge density σ on the cylinder which is equal to

σ = P(R) · r̂ = −kR.

Applying Gauss’s law on the cylinder:


Z
1 X
E · da = qi ,
0
i

we find that the normal component of the electric field will then be
σr ar
E⊥ = =−
0 0
from the range of 0 < r < R. We now use the fact that for a continuous distribution of charge,
the energy of the system is given as
Z Z
1 0
U= ρ(x)V (x)dV = |E|2 dV.
2 2
Therefore, the energy of the cylinder before it starts rotating with an angular velocity ω will be
2 R 2
π k 2 R4
Z  Z
0 kr 0 kr
U= dV = · 2πrdr = .
2 0 2 0 0 4 0

The volume current density at a point r = rr̂ + z ẑ of the cylinder is given as

J(r) = ρv = ρ(ω × r) = −2kωẑ × (rr̂ + z ẑ) = −2kωrθ̂.

To find the magnetic field of the cylinder, we can use Ampere’s Law:
∂B
∇ × B = µ0 J =⇒ − = −2µ0 kωr =⇒ B(r) = µ0 kωr2
∂r
where 0 < r < R. We use the fact that the energy due to the magnetic field is
Z
1
U= |B|2 dV
2µ0

1
Ashmit Dutta (June 2020) Day 266 Solution

to find that Z R
1 π
UB = (µ0 kωr2 )2 · 2πrdr = µ0 k 2 ω 2 R 6 .
2µ0 0 6
The electric field of the cylinder remains the same as before, so the energy of the cylinder after
it starts rotating is
π k 2 R4 π
U = UE + UB = + µ0 k 2 ω 2 R 6 .
4 0 6
There is extra energy because the cylinder has work done on it to make it moving at an angular
velocity ω. This extra work adds to the total energy of the system.

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