A coaxial cable consists of an inner cylindrical conductor of radius R 1 = 0.040

Question

A coaxial cable consists of an inner cylindrical conductor of radius R 1 = 0.040 m on the axis of an outer hollow cylindrical conductor of inner radius R 2 = 0.080 m and outer radius R 3 = 0.090 m. The inner conductor carries current I 1 = 1.3 A in one direction, and the outer conductor carries current I 2 = 7.8 A in the opposite direction.

A) What is the magnitude of the magnetic field at the following distances from the central axis of the cable? ( 0 = 4 × 10-7 T · m/A)

B)At r = 0.150 m (outside the cable)
At r = 0.060 m (in the gap midway between the two conductors)

Explanation / Answer

here,

R1 = 0.04 m

R2 = 0.08 m

R3 = 0.09 m

I1 = 1.3 A

I2 = 7.8 A

magnetic feild due to current carying wire , B = u0 * I /(2 * pi * d)

a)

at r = 0.15 m, outside the cable

magnetic feild ,B = u0 * I1 /(2 * pi * r) - u0 * I2 /(2 * pi * r)

B = u0 * 1.3 /(2 * pi * 0.15) - u0 * 7.8/(2 * pi * 1.5)

B = 8.67 * 10^-6 N/C

magnitude of magnetic feild at r = 0.15 is 8.67 * 10^-6 N/C

b)

at r = 0.060 m (in the gap midway between the two conductors)

as the magnetic field due to shell is zero inside the shell

the magnetic feild at r = 0.06 m , B = u0 * I1 /(2 * pi* r)

B = u0 * 1.3 /(2 * pi* 0.06)

B = 4.36 * 10^-6 N/C

the magnitude of the magnetic feild at r = 0.06 m is 4.36 * 10^-6 N/C

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