801_PartUniversity Physics Solution

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 5

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online