The figure below is a cross-sectional view of a coaxial cable. The center conduc

Question

The figure below is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.16 A out of the page and the current in the outer conductor is I2 = 2.96 A into the page. Assuming the distance d = 1.00 mm, answer the following.

(a) Determine the magnitude and direction of the magnetic field at point a.

(b) Determine the magnitude and direction of the magnetic field at point b.

magnitude ___________µT direction ---Select--- to the left, to the right, upward, downward, into the page, out of the page

Explanation / Answer

Here , as the magnetic field due side the current shell is zero

hence , for part a)

magnetic field at a ,

Ba = u0*I1/(2*pi*d)

Ba = 4pi*10^-7 * 1.16 /(2*pi*0.001)

Ba = 2.32 *10^-4 T

the magnetic field at the point a is the 2.32 *10^-4 T upwards

b)

Now, at the point b)

magnetic field at b m, Bb = B2 - B1

Bb = u0*I2/(2*pi*3d) - u0*I1/(2*pi*3d)

Bb = 4pi*10^-7 * 2.96 /(2*pi*0.003) - 4pi*10^-7 * 1.16 /(2*pi*0.003)

Bb = 1.2002 *10^-4 T

using right hand thumb rule

the magnetic field at point B is 1.2002 *10^-4 T into the page

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