Problem%20Set%205%20Solutions

Problem%20Set%205%20Solutions - 8 Letting F = q E v B = 0...

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8. Letting G G G G Fq EvB =+ × = di 0, we get sin vB E φ = . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin factor is at its maximum value (which is 1, corresponding to = 90°). So 3 3 min 1.50 10 V/m 3.75 10 m/s 0.400 T E v B × == = × .
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14. For a free charge q inside the metal strip with velocity G v we have G G G G Fq EvB =+ × di . We set this force equal to zero and use the relation between (uniform) electric field and potential difference. Thus, v E B VVd B xy x y == = × ×× = −− 390 10 120 10 0850 10 0 382 9 32 . .. V Tm ms ch c hc h
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26. Using 2 / Fm vr = (for the centripetal force) and 2 /2 Km v = , we can easily derive the relation K = 1 2 Fr . With the values given in the problem, we thus obtain K = 2.09 × 10 22 J.
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36. (a) Using Eq. 28-23 and Eq. 28-18, we find ( )( ) () 19 7 osc 27 1.60 10 C 1.20T 1.83 10 Hz. 2 21 . 6 71 0k g p qB f m × == = × π π× (b) From rm v q B m kq B pP 2 we have ( ) 2 19 2 7 27 19 0.500m 1.60 10 C 1.20T 1.72 10 eV. 2 2 1.67 10 kg 1.60 10 J eV p rqB K m −− ⎡⎤ × ⎣⎦ = × ××
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42. The magnetic force on the (straight) wire is () ( ) ( ) ( ) sin 13.0A 1.50T 1.80m sin 35.0 20.1N. B Fi B L θ == ° =
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axis. We take G B to be in the same direction as that of the current flow in the hypotenuse. Then, with BB == G 0 0750 .T , cos 0.0692T , sin 0.0288T. xy B B θ =− = = (a) Eq. 28-26 produces zero force when G G LB || so there is no force exerted on the hypotenuse of length 130 cm. (b) On the 50 cm side, the B x component produces a force iB yx A ± k, and there is no contribution from the B y component. Using SI units, the magnitude of the force on the A y side is therefore 4 00 0500 0 0692 0138 .. . . Am T N . bg b g b g = (c) On the 120 cm side, the B y component produces a force A ± k, and there is no contribution from the B x component. The magnitude of the force on the A x side is also 4 00 120 0 0288 . . T N . b g = (d) The net force is iB iB AA ±± , kk += 0 keeping in mind that B x < 0 due to our initial assumptions. If we had instead assumed G B went the opposite direction of the current flow in the hypotenuse, then B x > 0 but B y < 0 and a zero net force would still be the result. 48. We establish coordinates such that the two sides of the right triangle meet at the origin, and the A y = 50 cm side runs along the + y axis, while the A x = 120 cm side runs along the + x axis. The angle made by the hypotenuse (of length 130 cm) is = tan –1 (50/120) = 22.6°, relative to the 120 cm side. If one measures the angle counterclockwise from the + x direction, then the angle for the hypotenuse is 180° – 22.6° = +157°. Since we are only asked to find the magnitudes of the forces, we have the freedom to assume the current is
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Problem%20Set%205%20Solutions - 8 Letting F = q E v B = 0...

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