A coaxial cable has an inner conducting wire carrying current Ii = 5.0 A and an

Question

A coaxial cable has an inner conducting wire carrying current Ii = 5.0 A and an outer conducting layer carrying current Io = 10.0 A as shown below. The inner cable has a radius of 0.5 cm and the outer conducting layer has a radius of 1.0 cm. There is air in between the two conductors.

a) Using Ampere’s Law, determine the magnitude and direction of the magnetic field at a distance of 0.6 cm from the center.







b) Using Ampere’s Law, determine the magnitude and direction of the magnetic field at a distance of 1.2 cm from the center?

A coaxial cable has an inner conducting wire carrying current Ii = 5.0 A and an outer conducting layer carrying current Io = 10.0 A as shown below. The inner cable has a radius of 0.5 cm and the outer conducting layer has a radius of 1.0 cm. There is air in between the two conductors. a) Using Ampere's Law, determine the magnitude and direction of the magnetic field at a distance of 0.6 cm from the center. b) Using Ampere's Law, determine the magnitude and direction of the magnetic field at a distance of 1.2 cm from the center?

Explanation / Answer

Current through inner cable I1=5A Current through outer cable I2=10A Radius of inner cable R=0.5 cm current directions are   inner conducting wire has current direction into the page Outer conductor out of the page.      (a) magnetic field    B = 0I / 2r             Here r = 0.006 m Since r is inside for outer and outside for inner cable So outer cable is not contribute the field So outer cable is not contribute the field         I   =   current enclosed =    0.5A From amperes law field due to for inner coil at a distance r is given by         B=0I/2r      B =   ( 4*10-7 H /m)   ( 5 A) / (2 (0.006m )           =   1.67*10-4 T        Direction is clock wise direction (b)  Since r>R This point outsdie the two conductors But net currents due to both of them are in opposite directions IN=10A-5A=5A magnetic field    B = 0IN / 2r             Here r = 0.012 m           B =   ( 4*10-7 H /m)   ( 5A)   / (2 (0.012)              =   0.833 *10-4T           Direction is counter clock wise direction

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