Figure P30.23 is a cross-sectional view of a coaxial cable. The center conductor

Question

Figure P30.23 is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, which is surrounded by an outer conductor, which is surrounded by another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.00 A out of the monitor, and the current in the outer conductor is I2 = 3.04 A into the monitor. Determine the magnitude and direction of the magnetic field at point a.
µT   ---Direction--- to the left to the right upward downward into the monitor out of the monitor
Determine the magnitude and direction of the magnetic field at point b.
µT   ---Direction--- to the left to the right upward downward into the monitor out of the monitor

Explanation / Answer

At point a using

Ampere's law

B*dl = mu_o*I

mu_o is the permeability of free space

B*(2*pi*r1) = (4*3.142*10^-7*1) = 1.26*10^-6

B = 1.26*10^-6 /(2*3.142*1*10^-3)

B = 200*10^-6 T

B = 200 uT

direction is Upwards

b) at point b

using ampere's law

B*dl = (mu_o*I2)

B*2*pi*r2 = mu_o*(I2_i1)

B = (4*3.142*10^-7(3.04-1))/(2*3.142*3*10^-3)

B = 1.36*10^-4 T

B = 136*10^-6 T

B = 136 uT

direction is downwards

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