Tutorial 12 (week 13) - Solution T10.2 Two electrons (qe = - 1.6 x 10-19C, me = 9.11 x 10-31 kg) labelled A and B enter a uniform magnetic field of magnitude 1.5 T with different velocities. Within the field region, they follow curved paths confined to the xy-plane, then emerge at different locations on the y-axis, as shown. a) What is the direction of the magnetic field? Forces at points A and B need to be to the left. The velocity is upwards at these points. RHR gives into the page for positive charges, so out of the page for electrons. b) We know that the electrons have the same charge and the same mass. What is the ratio of their velocities, expressed in terms of a and b (see figure)? y From F = m a, we get for circular motion: qvB=mv2/r Uniform ? field Solving for v, we find: b B a A m v=(qBr) So the velocity is proportional to the radius. So VB > VA. q c) Let T be the amount of time each particle is in the shaded region that contains the magnetic field. How do the times TA and TB that each particle spend in the field compare? d) Write an expression for the time that electron A spends in the magnetic field. Use m, q, B and known constants like T. Solving part (d) first, we start with the equation for the velocity in part (b). Remember that v=As At Here, s is a half circle, so As = Tr At = Tis what we are solving for. Using the equation for velocity from part (b): T=mx q B We see that the time just depends on the strength of the field, so TA = TB.
T9.3 A capacitor was charged to 30 V and then discharged through a resistor. The figure shows the capacitor voltage as a function of time as the capacitor discharges through the resistor. a) What is the time constant for this RC circuit? b) If the capacitor in this circuit is a 26 uF-capacitor, what is the value of the resistance through which it is being discharged? c) How long would it take for the capacitor to discharge such that its charge is only 10% of the initial value? AVc (V) 30 Since V =, we know that the voltage follows the same decay curve as the charge: 20 V(t)=Voe+ We can get an exact result for the time constant by reading 10 a round value off the graph: The voltage falls from 30 V to 10 V in 4 ms: 0 10V=30 Ve -4 ms 1 0 2 -4 4 > t(ms) 6 1 -4 ms =e T 3 Taking the ln( ) on both sides: -In(3) == 4ms Solving for t gives T = RC = 3.64 ms. (b) Solving for R gives R = 140 ?. c) Now that we have the time constant, we can use V(t)=Voet again and solve for t: