chap26 - MAGNETISM: FORCE AND FIELD 26 EXERCISES Section...

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MAGNETISM: FORCE AND FIELD 26 EXERCISES Section 26.2 Magnetic Force and Field 17. INTERPRET This problem is about the magnetic force exerted on a moving electron. DEVELOP The magnetic force on a charge q moving with velocity v G is given by Equation 26.1: . B F qv B GG G The magnitude of B F G is ||| || | s i n BB FF q vBq v B θ == × = G G G EVALUATE (a) The magnetic field is a minimum when sin 1 = (the magnetic field perpendicular to the velocity). Thus, 15 3 min 19 7 5.4 10 N 1.61 10 T 16.1 G (1.6 10 C)(2.1 10 m/s) B F B qv × = ×× (b) For 45 , the magnetic field is 15 min 19 7 5.4 10 N 2 22.7 G ||s i n (1.6 10 C)(2.1 10 m/s)sin 45 B F × = = ° ASSESS The magnetic force on the electron is very tiny. The magnetic field required to produce this force can be compared to the Earth’s magnetic field, which is about 1 G. 18. (a) If the magnetic force is the only one of significance acting in this problem, then sin . Fm ae v B Thus, 31 15 2 19 5 v / sin (9.11 10 kg)(6 10 m/s )/(1.6 10 C)(0.1 T)sin 90 3.42 10 m/s. ma eB −− × × × ° = × (b) Since F vB × G z is perpendicular to the magnetic force on a charged particle changes its direction, but not its speed. , v G 19. INTERPRET In this problem we are asked to find the magnetic force exerted on a moving proton. DEVELOP The magnetic force on a charge q moving with velocity v G is given by Equation 26.1: . B F qv B G The magnitude of B F G is i n FFq v B q v B × = G G G The charge of the proton is 19 1.6 10 C. q EVALUATE (a) When 90 , the magnitude of the magnetic force is 19 5 14 sin 90 (1.6 10 C)(2.5 10 m/s)(0.5 T) 2.0 10 N B Fq v B = × × = × (b) When 30 , the force is 19 5 14 sin 30 (1.6 10 C)(2.5 10 m/s)(0.5 T)sin 30 1.0 10 N B v B = × × ° = × (c) When 0, the force is sin 0 0. B v B = ) ASSESS The magnetic force is a maximum ,max (| | B F qvB = when 90 = ° and a minimum when ,min (0 B F = ) 0. = ° 20. An electron, moving perpendicularly to the Earth’s magnetic field near the surface, experiences a maximum magnetic force of 19 4 3 19 31 v 2 / (1.6 10 C)(10 T) 2(10 eV)(1.6 10 J/eV)/(9.11 10 kg) eB eB Km × × .. 16 31 0 N . × The weight of an electron near the Earth’s surface is 31 2 30 (9.11 10 kg)(9.8 m/s ) 9 10 N, mg × . a factor of nearly times smaller. 14 0 × 21. INTERPRET This problem is about the speed of a given charge if it is to pass through the velocity selector undeflected. 26.1
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26.2 Chapter 26 DEVELOP In the presence of both electric and magnetic fields, the force on a moving charge is (see Equation 26.2): () EB F FFq EvB = += + × G GG G G G The condition for a charged particle to pass undeflected through the velocity selector is , E B F F =− or /. vE B = EVALUATE Substituting the values given in the problem statement, we obtain 24 kN/C 400 km/s 0.06 T E v B == = ASSESS Only particles with this speed would pass undeflected through the mutually perpendicular fields; at any other speed, particles would be deflected. Section 26.3 Charged Particles in Magnetic Fields 22. From Equation 26.3, the radius of the orbit is 27 19 2 / (1.67 10 kg)(15 km/s)/(1.6 10 C)(4 10 T) rm v e B −− = × × = 3.91 mm.(SI units and data for the proton are summarized in the appendices and inside front cover.) 23. INTERPRET This problem is about an electron undergoing circular motion in a uniform magnetic field. We want
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chap26 - MAGNETISM: FORCE AND FIELD 26 EXERCISES Section...

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