(a) A particle of mass m and electric charge e moves in a plane under the influe
ID: 1906352 • Letter: #
Question
(a) A particle of mass m and electric charge e moves in a plane under the influence ofa central force potential V( r) and a constant magnetic field B , perpendicular to the plane, generated by a static vector potential A=1/2 Bxr Find the Hamiltonian using coordinates in the observerExplanation / Answer
a) An electron moving in a magnetic field will experience a magnetic force. This magnetic force also acts as a centripetal force, i.e. F(cp) = F(mag) => mv2/r = Bqv . Cancel out the v's on both sides to get mv/r =Bq. Remember that angular frequency is defined as v/r => m(v/r) = Bq so, mw= Bq therefore w= Bq/m (1) b) w, angular frequency is also defined as 2pi/ T, where T=orbital period. So substituting this equation (1) we get 2pi/T = Bq/m . As it is the period that we are asked to calculate, rearrange the equation to get, T= 2(pi)m/ Bq. Now substitute the given values of B ((6.5x10^4) nT =6.5x10^-5) , the charge of an electron (e = 1.60 × 10^–19 C), mass of an electron (m = 9.11 × 10^–31 kg) into equation: T= 2(pi)m/ Bq.= 2(pi) (9.11 × 10^–31)/ (6.5x10^-5)(1.60 × 10^–19)= 5.5 x 10^-7 seconds c) Use the equation which i mentioned above, w= v/r and rearrange it to obtain r= v/w. Now substitute the given values into the equation to find the value of r ( radius of e's orbit) r= v/w = (3x10^5)/ w (REMEMBER w = 2pi/ T , so w= 2pi / (5.5 x 10^-7) = 1.14x 10^7 rad/s) r= (3x10^5)/ (1.14x 10^7) = 0.026 m = 26 mm d) Only look at the vertical component of the velocity of the electron, use Fleming's left hand rule to get the direction of the magnetic force acting on the vertical component of e's velocity... index finger= into paper, middle finger= points up, thumb = to the left => Direction of magnetic F on vertical component of v = to the left / direction of electric F is in z direction, use pythagoras theorem to find resultant F..... but not really sure what to do after that
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.