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Unformatted text preview: Physics 213 HW #7 Solutions Spring 2008 27.51. IDENTIFY: Drift velocity is related to current density by Eq.(25.4). The E field is determined by the requirement that the electric and magnetic forces on the current-carrying charges are equal in magnitude and opposite in direction. (a) SET UP: The section of the silver ribbon is sketched in Figure 27.51a. J x = n q v d so v d = J x n q Figure 27.51a EXECUTE: J x = I A = I y 1 z 1 = 120 A (0.23 10 3 m)(0.0118 m) = 4.42 10 7 A/m 2 v d = J x n q = 4.42 10 7 A/m 2 5.85 10 28 / m 3 ( ) 1.602 10 19 C ( ) = 4.7 10 3 m/s = 4.7 mm/s (b) magnitude of E : q E z = q v d B y , so E z = v d B y = (4.7 10 3 m/s)(0.95 T) = 4.5 10 3 V/m direction of E : The drift velocity of the electrons is in the opposite direction to the current, as shown in Figure 27.51b. v B F B = q v B = e v B Figure 27.51b The directions of the electric and magnetic forces on an electron in the ribbon are shown in Figure 27.51c. F E must oppose F B so F E is in the z-direction Figure 27.51c F E = q E = e E so E is opposite to the direction of F E and thus E is in the + z-direction. (c) The Hall emf is the potential difference between the two edges of the strip (at z = 0 and z = z 1 ) that results from the electric field calculated in part (b). E Hall = Ez 1 = (4.5 10 3 V/m)(0.0118 m) = 53 V EVALUATE: Even though the current is quite large the Hall emf is very small. Our calculated Hall emf is more than an order of magnitude larger than in Example 27.13. In this problem the magnetic field and current density are larger than in the example, and this leads to a larger Hall emf....
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