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A charged particle of mass m = 1 mg and charge q = 1 (mu) C enter along AB at point A in a uniform magnetic field B = 1.2 T existing in the rectangular region of size a xx b, where a = 4 m and b = 3 m. The particle leaves the region exactly at corner point C. What is the speed v (in m s^(-1)) of the particle? |
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Answer» `r=AO = CO = (mv).(QB) also a/r = sin theta, (r-B)/(r) = cos theta` Solve the above equaton to get  `v=qB(a^(2)+b^(2))/(2mb) = 10^(-6) xx 1.2 (4^(2)+3^(2))/(2 xx 10^(-6) xx 3) = 5 m//s`. |
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