Question

3. Refer to the diagram here (not to scale)-and note the coordinate axes. Note also L0 is castward and 290 is upward. This problem is about two different trials of the same basic experiment, using the apparatus shown here and two particles of equal mass m, but opposite.charge. Trial A: At time f 0, one particle, initially moving castward at speed v, enters the space between two charged plates of known length L, where a uniform electric field exists B Then the particle exits the plates and enters a uniform magnetic field B,, somewhere along that field's top edge. Then the particle exits B, (along the same top edge of the field), moving at velocity , as shown, and enters another uniform field, B, at time r. As it enters B a magnetic force F of known magnitude is exerted on the particle. Irial B: The above procedure is repeated for the other particle. All known data are the same, except the charge on the plates has been reversed. And in this trial, the particle reaches B, at time 1,. 234 cm The data kg 3.12 x 10 m/s225.0 B,89.0TL150 (All gravitational forces are negligible.) m1.50 x 10 B, 56.7 mT south F 7.28 x 10"N a. What was the difference in the total travel time in the two trials? That is, what was ,-2 b. Calculate the magnitude of the electric field being produced by the plates

All information required to solve the problem is provided in the question

Answer 1

Given that first we find the charge of the particle Using the data of the last part, we know F, V, B and the angle V & B there we can find the charge F = qVB sin (150 - 25 degree) q = F\v B sin (.125 degree) F 28 times 10^negative 18 Vertical-bar (3.12 times 10^5 m/s) (89 times 10^negative 4 +) (0.8192) q = 3.2 times 10^negative 21 C the sign of the chase should be (+)we assume that to wards the person reading the page as for the travel through the zone of B there is no info about the angle of the magnetic field in that area so, we can suppose that the magnetic field is perpendicular to the speed of the charge investment that event the particle inside said zone will have half a circumference as &the angle at the would be equal to the angle at the