In New England, the horizontal component of the earths magnetic fi eld has a magnitude of 1.6 3 1025 T. An electron is shot vertically straight up from the ground with a speed of 2.1 3 106 m/s. What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron
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Question
m A long, cylindrical conductor is solid throughout and has a radius R. Electric charges fl ow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in eff ect, a solid tube of current I0. The current per unit cross-sectional area (i.e., the current density) is I0/(pR2 ). Use Ampres law to show that the magnetic fi eld inside the conductor at a distance r from the axis is m0I0r/(2pR2 ). (Hint: For a closed path, use a circle of radius r perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)
Solution
The first step in solving 21 problem number 71 trying to solve the problem we have to refer to the textbook question: m A long, cylindrical conductor is solid throughout and has a radius R. Electric charges fl ow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in eff ect, a solid tube of current I0. The current per unit cross-sectional area (i.e., the current density) is I0/(pR2 ). Use Ampres law to show that the magnetic fi eld inside the conductor at a distance r from the axis is m0I0r/(2pR2 ). (Hint: For a closed path, use a circle of radius r perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)
From the textbook chapter Magnetic Forces and Magnetic Fields you will find a few key concepts needed to solve this.
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