801_PartUniversity Physics Solution

801_PartUniversity Physics Solution - 27-8 Chapter 27...

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27-8 Chapter 27 EVALUATE: Electromagnetic waves with frequency 3.0 THz f = have a wavelength in air of 4 3.0 10 m. v f λ == × The shorter the wavelength the greater the frequency and the greater the magnetic field that is required. B depends only on f and on the mass-to-charge ratio of the particle that moves in the circular path. 27.24. IDENTIFY: The magnetic force on the beam bends it through a quarter circle. SET UP: The distance that particles in the beam travel is s = R θ , and the radius of the quarter circle is R = mv / qB . EXECUTE: Solving for R gives R = s / θ = s /( π /2) = 1.18 cm/( π /2) = 0.751 cm. Solving for the magnetic field: B = mv / qR = (1.67 × 10 &27 kg)(1200 m/s)/[(1.60 × 10 &19 C)(0.00751 m)] = 1.67 × 10 &3 T EVALUATE: This field is about 10 times stronger than the Earth±s magnetic field, but much weaker than many laboratory fields. 27.25. IDENTIFY: When a particle of charge e is accelerated through a potential difference of magnitude V , it gains kinetic energy eV . When it moves in a circular path of radius R , its acceleration is 2 v R . SET UP: An electron has charge 19 1.60 10 C qe =− =− × and mass 31 9.11 10 kg × . EXECUTE: 2 1 2 mv eV = and 19 3 7 31 2 2(1.60 10 C)(2.00 10 V) 2.65 10 m/s 9.11 10 kg eV v m ×× == = × × . m = F a ! ! gives 2 sin v qvB m R φ = . 90 = ° and 31 7 4 19 (9.11 10 kg)(2.65 10 m/s) 8.38 10 T (1.60 10 C)(0.180 m) mv B qR = × × . EVALUATE: The smaller the radius of the circular path, the larger the magnitude of the magnetic field that is required. 27.26. IDENTIFY: After being accelerated through a potential difference V the ion has kinetic energy qV . The acceleration in the circular path is 2 /. vR SET UP: The ion has charge =+ . EXECUTE: . K qV eV + 2 1 2 mv eV = and 19 4 26 2 2(1.60 10 C)(220 V) 7.79 10 m/s. 1.16 10 kg eV v m × = × × sin . B Fq v B = 90 = ° . m = F a ! ! gives 2 v qvB m R = . 26 4 3 19 (1.16 10 kg)(7.79 10 m/s) 7.81 10 m 7.81 mm. (1.60 10 C)(0.723 T) mv R qB = × = × EVALUATE: The larger the accelerating voltage, the larger the speed of the particle and the larger the radius of its path in the magnetic field. 27.27. (a) IDENTIFY and SET UP: Eq.(27.4) gives the total force on the proton. At 0, t = () ²² ² ² . xz x z x qq v v B q v B =×= + × = F vB i k i j !! ! ( )( ) ( ) 19 5 14 1.60 10 C 2.00 10 m/s 0.500 T 1.60 10 N . −− × Fj j ! (b) Yes. The electric field exerts a force in the direction of the electric field, since the charge of the proton is positive and there is a component of acceleration in this direction. (c) EXECUTE: In the plane perpendicular to B ! (the yz -plane) the motion is circular. But there is a velocity component in the direction of , B ! so the motion is a helix. The electric field in the ² + i direction exerts a force in the ² + i direction. This force produces an acceleration in the ² + i direction and this causes the pitch of the helix to vary. The force does not affect the circular motion in the yz -plane, so the electric field does not affect the radius of the helix.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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801_PartUniversity Physics Solution - 27-8 Chapter 27...

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