Along, straight, solid cylinder, oriented with its axis in the z-direc

Agohofidov6 2021-12-16 Answered

Along, straight, solid cylinder, oriented with its axis in the z-direction, carries a current whose current density is J . The current density, although symmetric about the cylinder axis, is not constant and varies according to the relationship
J=(br)e(ra)/δk^ for r a
=0f or a
where the radius of the cylinder is a = 5.00 cm, ris the radial distance from the cylinder axis, b is a constant equal to 600 A/m, and δ is a constant equal to 2.50 cm.
(a) Let I0 be the total current passing through the entire cross section of the wire. Obtain an expression for I0 in terms of b, d, and a. Evaluate your expression to obtain a numerical value for I0.
(b) Using Ampere’s law, derive an expression for the magnetic field B in the region ra. Express your answer in terms of I0 rather than b.
(c) Obtain an expression for the current | contained in a circular cross section of radius r... a and centered at the cylinder axis. Express your answer in terms of I0 rather than b.
(d) Using Ampere’s law, derive an expression for the magnetic field B in the region ra.
(e) Evaluate the magnitude of the magnetic field at r- = 6, r= a, and r= 2a.

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Expert Answer

raefx88y
Answered 2021-12-17 Author has 26 answers

(a) Consider a long, solid cylinder which has a circular cross-section of radius a and oriented with its axis in the z- direction, the current density is not constant across the cross-section of the wire and it is given by
J=(br)e(ra)/δk^
where r is the distance from the center. First we need to find an expression for the total current passing through
the entire cross section of the wire I0 in terms of b,δ , and a. Consider a small segment, a ring with a width of dr and at distance from the center of the circle, the current in this element equals the area of the element multiplied by
the current density J, that is,
dI=JdA
the area of the element equals the area of a rectangle with a width of dr and length of 2πr (the circumference of a circle with a radius of r), that is,
dA=2πrdr
thus,
dI=(br)e(ra)/δ2πrdr=2πbe(ra)/δdr
the total current passing throught the wire equals the integration over the whole wire, from r = 0 to r = a, that is,
I0=0adI=2πb0aeraδdr
=2πbδeraδ{0}a=2πbδ(1eaδ)
I0=2πbδ(1eaδ
the radius of the cylinder is a=5.00,b is a constant equal to 600 A/m,and d is a constant equal to 2,50 cm, substitute with the givens to get,
I0=2π(600Am)(0.025m)(1e(0.0500.025))=81.5A
I0=81.5A

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Paul Mitchell
Answered 2021-12-18 Author has 40 answers

(b) Now for ra, the enclosed current is I0, therefore Ampere's law (B×dl=μ0Iencl) gives,
{B}×dl=B2πr=μ0Iencl=μ0I0
or,
B=μ0I02πr
(c) Now we need to find the enclosed current for ra. Consider a ring with width of dr and area of dA=2πdr
the enclosed current in this area is,
dl=JdA
or,
dI=2πberaδdr
integrate from r=0 to r=r to get,
dI=2πb0reraδdr=2πbδeraδ{0}r
=2πbδ(eraδeaδ)=2πbδeaδ(erδ1)
but I0=2πbδ(1eaδ),thus
I=I0(erδ1)(eaδ1)

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nick1337
Answered 2021-12-27 Author has 575 answers

(d) Now we need to use the results of part (c) to find the magnetic field for ra. Ampere's law (B×dl=μ0Iencl), gives,
B×dl=B(r)2πr=μ0Iencl=μ0I0(er/δ1)(ea/δ1)
or,
B=μ0I0(er/δ1)2πr(ea/δ1)
(e) Finally we need to evaluate the magnitude of the magnetic field at r=δ,r=a, and r=2a, as,
B(r=δ)=μ0I0(e1)2πδ(ea/δ1)
=(4π×107T×m/A)(81.5A)2π(0.025m)(e1)(e0.050/0.0251)
=1.75×104T
B(r=δ)=1.75×104T
and
B(r=a)=μ0I0(ea/δ1)2πa(ea/δ1)
=(4π×107T×m/A)(81.5A)2π(0.050m)
=3.26×104T
B(r=δ)=3.26×104T
and
B(r=2a)=μ0I02πa
=(4π×107T×m/A)(81.5A)2π(0.100m)
=1.63×104T
B(r=2a)=1.63×104T

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