Question

Refer to the diagram here (not to scale)-and note the cOordinate axes Note also: 20° is eastward and 290° is upward. east A particle of known mass m, initially moving eastward at an unknown speed vo, enters the space between two charged plates, of known length L, where a uniform electric field exists After exiting the plates, the particle enters, then exits a uniform magnetic field of known magnitude B. Then it enters the known uniform magnetic field B, at a known velocity v, As it enters B, a known magnetic force F is exerted on the particle. B B B 89.0 pTZ150 (All gravitational forces are negligible.) L 23.4 cm F =7.28x 10- N north The data: m 1.50 x 1030 kg =3.12 x 10 m/s Z 25.0 B, = 56.7 mT a. Describe the particle's travel within B At what angle did it exit? At what angle did it enter? How far apart were its entry and exit points? Calculate the total time the particle spent within B (There are two possible paths. Draw both, but find the transit time for just one-just pick one.) b. Calculate the electric field (magnitude and direction) produced by the plates c. Suppose that-instead of entering the plates or doing any of the other travel described above-this particle had just passed (at the velocity ,, same as above) through the center of a simple circular loop (R 5.46 cm) carrying a known steady current (7.89 A). If the force exerted on the particle as it passed through the center of the loop was the same as above (F 7.28 x 10-18 N north), in what direction was the loop's normal vector pointing? |(There are four possible solutions. Draw all four but calculate just one-just pick one.)

Answer 1

Given data: - m = 1.50 times 10^-20 kg h = 23.4 cm B_1 = 56.7 mu T B_2 = 89.0 mu T < 150 degrees V_2 = 3.12 times 10^5 m/s < 25.0 degrees F = 7.28 times 10^-18 N(north) using the data of the last part, we know, F, V, B and the angle between V and B, therefore we can find the charge. F = qvB sin(150 degrees - 25 degrees) q = F/v sin(125 degrees) q = 7.28 times 10^-8/(3.12 times 10^5)(89 times 10^-4 T)(0.8192) therefore q = 3.2 times 10^-21 C The sign of the charge should be (+) is we assume that 'north' is towards the person leading the page, As for the travel through the zone of B_1, there is no information 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. \r\n The particle inside said