Question

3. (a) Show that when a particle with mass m and charge q enters a magnetic field having its velocity v perpendicular to the direction of the magnetic field B, it will perform a mv circular path of radius R- qB (b) Using the previous result find an expression for the period T of the circular motion. (c) A charged particle moves into a region of uniform magnetic field, goes through half a circle and then exits that region, as in the diagram below. The particle is either a proton or an electron and it spends 100 ns within the region. Is the particle a proton or an electron (d) What is the magnitude of the magnetic field B? (e) If the same particle but with half the speed, enters again in the same magnetic field along the original direction, how long it will spend in the field (explain your answer) ?

Answer 1

(a) Magnetic force = qvB magnetic force = centripetal force qvB = mv^2/R R = mv/qB (b) Time period = pi r/v = pi times mv/qBv T = pi m/qB (c) F_B = q [v vector times B vector] v rightarrow -j hat B rightarrow k hat i hat leftarrow F_B = q[-j hat times k hat) = q/T [-i hat[-ve F_B rightwardsdoublearrow + i q rightarrow electron (d) T = 100 times 10^-9 s T = pi m_e/qb 100 times 10^-9 = pi times 9.11 times 10^-31/1.6 times 10^-19 times B B = 1.79 times 10^-4 T (e) T = pi m/qB doesn't depend on speed. T = 100 ns So time will remove same.