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A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is manitained in a direction normal to the common axis of the coils. A narrwo beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostaic field. If the beam remains undeflected when the electrostatic field is 9.0 xx 10^(-5) V m^(-1), make a simple guess as to what the beam contains. Why is the answer not unique? |
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Answer» Solution :Let charged be having a charge q, mass m and velocity ACQUIRED by them, when accelerated through a potential DIFFERENCE V, be v, then `v = sqrt((2 q V)/(m))` For the beam to remain undeflected `vecF_(E) and vecF_B` must be equal in magnitude but mutually opposite in direction. It is possible when `qE = q v B "or" v = E/B` `implies sqrt((2qV)/(m)) = E/B "or" q/m = (E^2)/(2 B^2 V)` In present case `E = 9.0 xx 10^(-5) V m^(-1) , B = 0.75 T and V = 15 kV = 15 xx 10^3 V` `:."" q/m = ((9.0 xx 10^5)^2)/(2 xx (0.75)^(2) xx 15 xx 10^(3)) = 4.8 xx 10^(7) C kg^(-1)` A simple guess suggests that HTE beam may contain deutron particle having mass `m = 3.34 xx 10^(-27) kg`and charge `q = +e = 1.6 xx 10^(-19)C`, thus , having a value of `q/m = (1.6 xx 10^(-19))/(3.34 xx 10^(-27)) = 4.8 xx 10^(7) C kg^(-1)` However , this answer is not UNIQUE because value of the ratio `q/m` may be `4.8 xx 10^(7) C kg^(-1)` for other ions like `Li^(+++), He^(++)` too. |
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