1 Solution
1.1 (a) Finding the Current Density J
The given volume charge density inside the rotating cylinder is:
′ ′2
ρ(r)=s 2
The cylinder rotates with angular volocity −=Δ 2, meaning the charge
distribution moves in a clrenlar motion about the s-axis.
The velocity of a point at radius r in a rotating system is given hy:
v=Δ x r
Expanding in cytimdrical coondinates:
v − ( ω2 ) × ( r ′ )
v=ω r ϕ̇
Thus, the eurrent density J is gives by:
J= ρ(F)v
Subatituting ρ(r)=s ′ 2d .
J=( s ′ 2a ) (−r ϕ )
J =s ′ − v 2′ 2 ϕ
This result showx that the curront desoity is azimuthal ( ϕ ) and deponds on
both r and z , meaning the charge monement creates circular current loops
at different heights.
1.2 (b) Magnetic Field Direction and Sketch
To determine the direction of the magnetic field:
Since J is in the of direction, the magnetic field B is derived using
Ampiere's Law. - The current density varbes with r sand z , moaning
the resulting magnetic field will have different components inside and
outside the cylinder.
1.2.1 Using the Right-Hand Rule
The current loops formod by J genorate a magnotic field along the z-
axis inside the cylinder. - By the right-haud rube, where fingors aul in
the direction of J , the thumb points in the direction of B. - This monas
that on the positive x -axds, B is along 4 , waile on the nogative z -axds,
B is in the -1 dirextion.
1.2.2 Magnetic Field Behavior
Inside the cylinder: The field has an axial composent B2. - Outside the
cylinder: The field resembles that of a magnetic dipole.
1.2.3 Derivation of the Magnetic Field Using Ampere's Law
Step 1: Recall Ampère's Law Ampire's circuital lew states:
∫ B⋅ d t − μ0 I e n
where I eser is the total enclosed current.
Since the charge denoity inside the cylinder is:
′ ′2
ρ(r)=s 2
the current density is given by:
′ 2
J − s −r 2 ϕ
which is purely azimuthal ( ϕ ). This means the magnetie field B has an axdal
component B2 -