A very long coaxial cable consists of a solid cylindrical inner conductor of rad
ID: 1390588 • Letter: A
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 Ampre'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 .
1.9410?4
Use Ampre'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.
Comment
B =1.9410?4
T I GOT THIS PART CORRECTExplanation / Answer
Radius of solid core r1 = 3.8 mm = 3.8x10-3 m
inner radius of cylindrical shell r2 = 7 mm = 7x10-3 m
outer radius of cylindrical shell r3 = 9.3 mm =9.3x10-3 m
current in the core I1 = 4.65 A
current in shell I2 = 4.6 A
a) Magnetic field between core and shell at r = 4.8 mm
Ampere's law, 2pi r B = muo I1
B= muo x I1 /( 2 pi r)
= 1.257x10-4 x 4.65 / (2 x 3.14 x 4.8x10-3)
B= 1.939x10-4 T
b) Magnetic field at r= 7.1 mm in shell lies between r2 < r < r3
current through the cross sectional area of the shell Ie = integal of ( J A)
= J integral of |r2r( pi r2 dr) |02pi d(theeta)
Ie = 2 pi J (r3 - r23) /3
When, r = r3, Ie = I2
I2 = 2 pi J (r33 - r23)/3
J = 3I2 / (2 pi ( r33 - r23)
therefore Ie = (r3 - r23) I2 / (r33 -r23)
Net current in the system I = -I1 + Ie where into the page is negative, out of page is positive
Magnetic field at point r= 7.1mm
2 pi r B = muo I
B= muo I / (2 pi r)
=(mu0/ 2 pi r ) { -I1 + ( r3 - r23) I2/ (r33- r23)}
= (1.257x10-6 / 2 x 3.14 x 7.1x10-3) { -4.65 + 4.6 ( 7.1x10-3^3 - 7x10-3^3)/ (9.3x10-3^3 - 7x10-3^3)
= 2.819x10-5 ( - 4.65 + 3.78x10-4)
= 2.819x10-5 x (- 4.6496)
B = -1.3107x10-4 T
The negative sign indicates magnetic field is into the page.