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Ampheres law

A long metal cylinder of radius (a) has the z-axis as its axis of symmetry. The cylinder carries a steady current of uniform current density
(Vector)J=(J subscript (z)) ((vector)e subscript(z)).
Derive an expression for the magnetic field at distance r from the axis where r<a.
By resolving the cylindrical unit (vector)e subscript (phi) along the x and y axes, show that the magnetic field at any point P inside the cylinder is
vectorB(x,y,z)=( ((miu) (subscript o))/2 )J subscript z (-ye subscript x +x (vector)e(subscript)y)
where P has the Cartesian coordinates (x,y,z) and (x^2 +y^2)<a^2.

Part B
A cylindrical hole of radius b,a is drilled through the cylinder.The axis of the hole is parallel to the axis of the cylinder,but is displaced from it by a distance d in the x-direction where d<a-b. The asymmetric hollow cylinder has a uniform current density
vectorJ=J subscript z *vector e subscript z
throughout the remaining material of the cylinder, and there is no current in the hole.
Use the result of part (a) to show that the magnetic field inside the cylindrical hole is uniform by the expression B=((miu) subscript 0)/2*(J subscript z)* d*(vector e subscript y )

Solution Preview

Please see the attached.

Reading: Right hand rule
The magnetic field due to a long straight wire is haveng circular lines the direction is given by right hand rule.
According to this rule if we emagine to hold the wire with our right hand such that thumb is showing the direction of current in the wire, then the direction of megnetic feld is given by the curled fingures.
In case a) the current is inward so the thumb of right hand should point inward then the fingures points clockwise direction. Hence the magnetic field is in clockwise direction.
Similarly the in case b) the current is outward and keeping the thumb upward the direction of the magnetic field lines will be anticlockwise.

A long metal cylinder of radius (a) has the z-axis as its axis of symmetry. The cylinder carries a steady current of uniform current density (Vector)J=(J subscript (z)) ((vector)e subscript(z)).
Derive an expression for the magnetic field at distance r from the axis where r<a.
By resolving the cylindrical unit (vector)e subscript (phi) along the x and y axes, show that the magnetic field at any point P inside the cylinder is vector B(x,y,z)=( ((miu) (subscript o))/2 )J subscript z (-ye subscript x + x ...

Solution Summary

The expert derives an expression for the magnetic field at a distance.

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