Question

ats each) For each of the following problems, a particle with charge q is traveling with velocity through a region ugh a region with constant magnetic field B. Write the force on the particle Fs in terms of its x, y, and components. In order to help you understand the cross product, I suggest that you solve all of these roblems using the vector cross product then confirm the result using the magnitude of the force and right hand rule for the direction. (a) 7.20 AC, U = 9.30 m/s, and B -11.31 mm (0 i + (0)+ (757) k) nN (b) q- 6.60 MC, 0 = 1.60 i m/s, and B = 3.80k mT (0 i + (-10.1) 3 + (0) k) nN (e) = -3.2 C, v = -4.2 k m/s, and B = -3.2k mT (0 i + (0) ) + (0) k) nN (+) mT (d) = -5.5 MC, 0 = -10.8 k m/s, and B = 10.6 (-445 i + (445) 3 + (0) ) nN (e) = 9.30 C, v = -2.7 i m/s, and B = -9.5 (+3) mT (0i + (0) ) + (169) k) nN

Answer 1

(a) q = 7.2046 = 9.30 } mis and B = -11.3 i mł F = 9ūXB) = (7.2044).[9:30 31x61.3)]x1ő3 = 7.20 x100 [105.09 9 + i)] 103 = 756.648 xlo? êxj) = 7574109 ľ [:?x} =k] F = 757 Ř MN = (oitoj +7576) AN. (Along the z-axis) (b) q = 6.60 MC. = 1.60 î m/b and B = 3.80k mT. = q (u x B) = (6 60x106). R1.60 ?)X(3.80 Ŕ)7x103 = 40-128x109 lxk) = - 40.1 ŷ nn. F = col - 40.1 h + od nn | (Along ve y-axis) () q = -3.2 MC, J = -4,2 mis; and © = - 3.2Âm T. . = q (xB') or 191(TxD) = 3.2x106. [-4.2).(-3-2) CRXK)]x1013 : Ě = 0 (kxk =o) 1 ²=coitojtoh) w N. I (d) 191 = 5.5 MC, ū==10.8 Â mis and ŏ = 1016.cit) mł. .: 72191 CTXB) = (515x156).[(-10.8k) x 1016. City)]xro = 445. 28x10 ? [~² x û +} ]] = 445 (- R x ² = { x8, RN = 445 (-; tî Inn F = [445 1 + (-445) } to RJ nN /

(e) q = 9.3046 , P = -2.71 mis and t = -9.5 t cithimT . Fzq (2 x B) = (9.30x106) [-2.72 ) x. (- 9.5 cît ? )] 103 = 1687x109 [ix² + x ÎN = 169 (0.t inn P = (ol toi + 1692) na Abony z-axis.