hw6 - Problem 2 An electron moving at right angles to a...

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Problem 2. An electron moving at right angles to a 0.10-T magnetic field experiences an acceleration of 6.0 × 10 15 m/s 2 . (a) What is the electron’s speed? (b) By how much does its speed change in 1 ns ( = 10 9 s) ? Solution (a) If the magnetic force is the only one of significance acting in this problem, then F = ma = e v B sin θ . Thus, v = ma = eB sin θ = ( 9 . 11 × 10 31 kg )( 6 × 10 15 m/s 2 ) = ( 1 . 6 × 10 19 C )( 0 . 1 T )sin 90 ° = 3 . 42 × 10 5 m/s. (b) Since F ª v × B is perpendicular to v , the magnetic force on a charged particle changes its direction, but not its speed. Problem 5. A particle carrying a 50- μ C charge moves with velocity v = 5 . 0 î + 3 . 2 ˆ k m/s through a uniform magnetic field B = 9 . 4 î + 6 . 7 ˆ j T. (a) What is the force on the particle? (b) Form the dot products F v and F B to show explicitly that the force is perpendicular to both v and B . Solution (a) From Equation 27-2, F = q v × B = ( 50 μ C )( 5 î + 3 . 2 ˆ k m/s ) × ( 9 . 4 î + 6 . 7 ˆ j T ) = ( 50 × 10 6 N )( 5 × 6 . 7 ˆ k + 3 . 2 × 9 . 4 ˆ j 3 . 2 × 6 . 7 î ) = ( 1 . 072 î + 1 . 504 ˆ j + 1 . 675 ˆ k ) × 10 3 N . (The magnitude and direction can be found from the components, if desired.) (b) The dot products F v and F B are, respectively, proportional to ( 1.072)(5) + (1.675)(3.2) = 0, and ( 1.072)(9.4) + (1.504)(6.7) = 0, since the cross product of two vectors is perpendicular to each factor. (We did not round off the components of F , so that the vanishing of the dot products could be exactly confirmed.) Problem 12. A velocity selector uses a 60-mT magnetic field and a 24 kN/C electric field. At what speed will charged particles pass through the selector undeflected? Solution The condition for zero deflection is v = E = B = (24 kN/C) = (0.06 T) = 400 km/s. Problem 13. A region contains an electric field E = 7 . 4 î + 2 . 8 ˆ j kN / C and a magnetic field B = 15 ˆ j + 36 ˆ k mT. Find the electromagnetic force on (a) a stationary proton, (b) an electron moving with velocity v = 6.1 î Mm/s. Solution The force on a moving charge is given by Equation 29-2 (called the Lorentz force) F = q ( E + v × B ). (a) For a stationary proton, q = e and v = 0 , so F = e E = ( 1 . 6 × 10 19 C )( 7 . 4 î + 2 . 8 ˆ j ) kN/C = (1.18 î + 0 . 448 ˆ j ) fN. (b) For the electron, q = e and v =
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