Chapter 5: Q44P (page 258)
Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.
Short Answer
The magnetic force of attraction is .
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Chapter 5: Q44P (page 258)
Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell.
The magnetic force of attraction is .
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A current flows to the right through a rectangular bar of conducting material, in the presence of a uniform magnetic fieldpointing out of the page (Fig. 5.56).
(a) If the moving charges are positive, in which direction are they deflected by the magnetic field? This deflection results in an accumulation of charge on the upper and lower surfaces of the bar, which in turn produces an electric force to counteract the magnetic one. Equilibrium occurs when the two exactly cancel. (This phenomenon is known as the Hall effect.)
(b) Find the resulting potential difference (the Hall voltage) between the top and bottom of the bar, in terms of,(the speed of the charges), and the relevant dimensions of the bar.
(c) How would your analysis change if the moving charges were negative? [The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material.]

A circular loop of wire, with radius , R lies in the xy plane (centered at the origin) and carries a current running counterclockwise as viewed from the positive z axis.
(a) What is its magnetic dipole moment?
(b) What is the (approximate) magnetic field at points far from the origin?
(c) Show that, for points on the z axis, your answer is consistent with the exact field (Ex. 5.6), when .
(a) By whatever means you can think of (short of looking it up), find the vector potential a distance from an infinite straight wire carrying a current . Check that and .
(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.
(a) Prove that the average magnetic field, over a sphere of radius R,due to steadycurrents inside the sphere, is
whereis the total dipole moment of the sphere. Contrast the electrostatic
result, Eq. 3.105. [This is tough, so I'll give you a start:
Writeas ,and apply Prob. 1.61(b). Now put in Eq. 5.65, and do the
surface integral first, showing that
(b) Show that the average magnetic field due to steady currents outsidethe sphere
is the same as the field they produce at the center.
Question: Using Eq. 5.88, calculate the average magnetic field of a dipole over
a sphere of radius Rcentered at the origin. Do the angular integrals first. Compare your answer with the general theorem in Prob. 5.59. Explain the discrepancy, and indicate how Eq. 5.89 can be corrected to resolve the ambiguity at . (If you get stuck, refer to Prob. 3.48.) Evidently the truefield of a magnetic dipole is
Compare the electrostatic analog, Eq. 3.106.
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