(a)
Find the expression for magnitudes of the magnetic field in each of the four layers.
(a)
Answer to Problem 27PQ
The expression for magnitudes of the magnetic field in each of the four layers are
Explanation of Solution
Write the expression for Ampere’s Law for the area bounded by the curve as.
Rearrange above equation for
Here,
Write the expression for current density as.
Here,
Write the expression for area of the circle as.
Substitute
Consider that the inner coaxial cable of radius
Write the expression for Amperian loop current as.
Substitute
Write the expression for Amperian loop is the circumferences of the circle as.
Substitute
Rearrange the above equation as.
Thus, the magnitude of the magnetic field in inner layer
Consider the inside insulator coaxial
The current in the loop is product of current density and area of loop is
The length of Ameprian loop is equal to the circumferences of the circle that is
Substitute
Thus, the magnitude of the magnetic field in inside insulator layer is
Consider the outer coaxial of radius
Write the current density of the layer
Substitute
Write the expression for Amperian loop current as.
Substitute
Substitute
Here,
Thus, the magnetic field on the third layer is
Write the expression for net magnetic field at the layer
Substitute
The net current enclosed inside the coaxial cable is zero.
Thus, the magnitude of the magnetic field outside the insulator layer is zero that is
Conclusion:
Thus, the expression for magnitudes of the magnetic field in each of the four layers are
(b)
Compare the part a results with the magnetic field produced by a long, straight wire and explain the advantage of using a coax.
(b)
Answer to Problem 27PQ
The expression for magnitude of the magnetic field for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.
Explanation of Solution
Write the expression for the magnetic field strength (magnitude) produced by a long straight current-carrying wire as.
The expression for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.
The advantage of using of coaxial cable as:
- 1. The inner conductor is in a Faraday shield, noise immunity is improved, and coax has lower error rates and therefore slightly better performance than twisted-pair.
- 2. Coax provides sufficient frequency range to support multiple channel, which allows for much greater throughput.
- 3. It also provides greater spacing between amplifiers coax's cable shielding reduces noise and crosstalk
Conclusion:
Thus, the expression for magnitude of the magnetic field for long straight current carrying wire is equal to the magnitude of the magnetic field in inside insulator layer.
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Chapter 31 Solutions
Physics for Scientists and Engineers: Foundations and Connections
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- A metal rod of mass M and length L is pivoted about a hinge at point O as shown in Figure P32.80. The axis of rotation passes through O into the page. A constant magnetic field B is applied into the page. Find the ratio of the maximum electric field inside the rod to the applied magnetic field when the rod is rotated with angular speed . Assume the speed of the rod is determined by the linear speed of its center of mass, and its mass is uniformly distributed. FIGURE P32.80arrow_forwardA uniform magnetic field B=5.44104iT passes through a closed surface with a slanted top as shown in Figure P31.59. a. Given the dimensions and orientation of the closed surface shown, what is the magnetic flux through the slanted top of the surface? b. What is the net magnetic flux through the entire closed surface?arrow_forwardFigure P30.11 shows three configurations of wires and the resultant magnetic fields due to current in the wires. What is the direction of the current that gives the resultant magnetic field shown in each case?arrow_forward
- A Derive an expression for the magnetic field produced at point P due to the current-carrying wire shown in Figure P30.26. The curved parts of the wire are pieces of concentric circles. Point P is at their center.arrow_forwardA current-carrying conductor PQ of mass m and length L is placed on an inclined plane with angle of inclination (Fig. P30.93). A uniform magnetic field B is directed upward as shown. Assume friction is negligible. a. Determine the magnitude and direction of the current in the conductor so that it remains in equilibrium. b. If the direction of the current is reversed, will the conductor still be in equilibrium? If not, find the magnitude of the initial acceleration of the conductor. FIGURE P30.93arrow_forwardThe magnetic field in the certain region is 0.128 T, and its direction is that of the +z-axis in the figure below. (a) What is the magnetic flux across the surface abcd in the figure? (b) What is the magnetic flux across the surface befc? (c) What is the magnetic flux across the surface aefd? (d) What is the net flux through all five surfaces that enclose the shaded volume?arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning