hw16 - Physics 21 Fall, 2010 Solution to HW-16 28.48 The...

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Unformatted text preview: Physics 21 Fall, 2010 Solution to HW-16 28.48 The current in the windings of a toroidal solenoid is 2.400 A. There are N = 500 turns and the mean radius is r = 25 . 00 cm. The toroidal solenoid is filled with a magnetic material. The magnetic field inside the windings is found to be 1.940 T. (a) Calculate the relative permeability. (b) Calculate the magnetic susceptibility of the material that fills the toroid. (a) The magnetic field inside a tightly wound toroidal solenoid is B = K m nI = K m NI 2 r , where n is the number of turns per unit length and N is the total number of turns. Solving the last equation for K m , we get K m = 2 rB NI = 2 . 25 1 . 94 (4 10- 7 ) 500 2 . 4 = 2021 . (b) The magnetic susceptibility is m = K m 1 thus the answer is 2020. 28-52 A long, straight wire carries a current of 2.50 A. An electron is traveling in the vicinity of the wire. At the instant when the electron is 4.50 cm from the wire and traveling with a speed of 6 . 00 10 4 m/s directly toward the wire, what are the magnitude and direction (relative to the direction of the current) of the force that the magnetic field of the current exerts on the electron? I v B F The magnetic field due to the wire has magnitude B = I 2 r and direction (from the rh rule) out of the page at the loca- tion of the electron as shown. The force on the electron is given by F = q v B = e v B . F is in the direction shown in the diagram. Note that the negative sign of the electron makes F in the opposite direc- tion from v B . The vectors F , v , and B are mutually perpendicular, so the magnitude of F is F = evB = ev I 2 r Using SI units, F = (1 . 602 10- 19 )(60000)(4 10- 7 )(2 . 5) 2 ( . 045) N = 1 . 07 10- 19 N The magnetic field exerts a force in the same direction as...
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hw16 - Physics 21 Fall, 2010 Solution to HW-16 28.48 The...

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