Question

A very long coaxial cable consists of a solid cylindrical inner conductor of radius 3.8mm , surrounded by an outer cylindrical conductor with inner radius 7mm , and outer radius 9.3mm . The region between the two conductors is filled with a waxlike insulating material to keep the conductors from touching each other. The inner conductor carries a current 4.65A , into the page, while the outer conductor carries a current 4.6A , out of the page.

Part A

Use Amp�re's law to derive an expression for the magnetic field as a function of r(the distance from the central axis) at points between the two conductors (R1<r<R2i). Use this expression to find the magnetic field at 4.8mm .

B =

1.94�10^(-4)

T [I GOT THIS PART CORRECT]

Use Amp�re's law to derive an expression for the magnetic field as a function of r(the distance from the central axis) at points inside the outer conductor (R2i<r<R2o). Use this expression to find the magnetic field at 7.1mm .

Use into page as negative and out of page as positive current.

B =		T

Comment

Hint 1. Ampere's Law

Ampere's Law states:  ?B??s=?0?Ienclosed Therefore you need to figure out what the enclosed current is for this case.

Hint 2. Enclosed current

The enclosed current is the current in the inner conductor + that portion of the current enclosed in the outer conductor. Use Itotal/Atotal=Ienclosed/Aenclosed Don't forget Atotal=pi?(R22o?R22i) and there is a corresponding expression for Aenclosed.

Answer 1

for outer cylindrical conductor ::

R_i = 7 mm = 7 x 10^-3 m

R_o = 9.3 mm = 9.3 x 10^-3 m

Area of cross-section of outer cynlindrical conductor is given as ::

A_outer = (R_i² - R²_o)

A_outer = (3.14) ((9.3 x 10^-3 )² -(7 x 10^-3 )²)

A_outer = = 1.18 x 10^-4 m²

total current in outer conductor = I_o = 4.6 A

current density in outer conductor = I_o/A_outer = 4.6 / (1.18 x 10^-4 ) = 3.89 x 10⁴ A/m²

current enclosed in the assumed circle of radius ''r'' in outer condutor is given as

I = current density x Area of assumed circle

I = 3.89 x 10⁴ ((3.14) ((r )² - (7 x 10^-3 )²))

I = 12.215 x 10⁴ ((r )² - (7 x 10^-3 )²))   outward

current in inner conductor = I_inner = 4.65 A inward

net current enclosed, I_enclosed = I - Inner = 12.215 x 10⁴ ((r )² - (7 x 10^-3 )²)) - 4.65 A

Using ampere's law

B. (2r) = u_oI_enclosed

B = u_oI_enclosed / (2r)

B = u_o (2I_enclosed )/ (4r)

B = u_o (2) (12.215 x 10⁴ ((r )² - (7 x 10^-3 )²)) - 4.65 A )/ (4r)               Eq-1

since r = 7.1 mm = 7.1 x 10^-3 m

B = (2x 10^-7) (12.215 x 10⁴ ((7.1 x 10^-3)² - (7 x 10^-3 )²)) - 4.65 A )/ (7.1 x 10^-3)    (u_o/4 = 10^-7 )

B = -1.26 x 10^-4 T