Refer to the diagram here (not to scale) and note the coordinate axes. A particl
ID: 1413543 • Letter: R
Question
Refer to the diagram here (not to scale) and note the coordinate axes. A particle of known mass m, initially moving eastward at an unknown speed v_0, 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_1. Then it enters the known uniform magnetic field B_2 at a known velocity v_2. As it enters B_2, a known magnetic force F is exerted on the particle. The data: m = 1.50 Times 10^-26 kg L = 23.4cm B_1 = 56.7 mT B_2 = 89.0 mu T 150 degree v_2 = 3.12 Times 10^5 m/s 25.0 degree F = 7.28 Times 10^-18 N north (All gravitational forces are negligible.) Describe the particle's travel within B_1: 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_1. (There are two possible paths. Draw both, but find the transit time for just one just pick one.) Calculate the electric field (magnitude and direction) produced by the plates. Suppose that instead of entering the plates or doing any of the other travel described above this particle had just passed (at the velocity v_0, same as above) through the center of a simple circular loop (R = 5.46 cm) carrying a known steady current (I = 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 Times 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.)Explanation / Answer
First we find the charge of the particle:
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 = qvBsin(150°-25°) :::: q = F/vBsin(125°) :::::: q = 7.28*10-18 /(3.12*105 m/s)(89*10-4 T)(0.8192)
q = 3.2*10-21 C
The sign of the charge should be (+) if we assume that 'north' is towards the person reading the page.
As for the travel througth the zone of B1, 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. In that event, the particle inside said zone will have half a circumference as trajectory and the angle at the exit would be equal to the angle at the entry, but with opposite sign. So:
a. Angle at the exit: (according to the data) 25° above the horizontal
Angle at the entry: 25° below the horizontal
Distance between said points (two times the radius of the circumference):
r = mv2/qB1 = (1.50*10-26 kg)(3.12*105 m/s) / (56.7*10-4 T)(3.2*10-21 C) = 258 m
So, the distance between the points is: D = 516 m
Note: we have got a very big number as an answer for this question, this could be due to either the big mass of the particle (usually is something*10-27) and its small charge (usually something*10-19) or because the supposition made about the direction of B1 was not correct.
The time of the particle inside that zone would be half of the period of the circular movement that it describes, so:
T = 2/w ; where w = qB/m ::::::: w = (3.2*10-21 C)(56.7*10-4 T)/(1.5*10-26 kg) = 1210 rad/s
T = 5.2*10-3 s :::::::: So, the time inside the zone would be t = (1/2) T :::::::: t = 2.6*10-3 s
b. The magnitude of the of the electric field can be determined if we know the electric force, which can be found once we know the acceleration of the particle.
To find the acceleration we can assume that the electric field goes downwards and therefore so does the acceleration. If that is the case, then we can use the following relations from kinematics to solve this issue:
v2 = 3.12*105 m/s :::::: v2y = 1.32*105 -j m/s ; v2x = 2.83*105 i m/s (the angle is -25°)
The distance travelled in the x direction is:
L = v2x*t :::::: t = L/v2x = L/vo = 23.4*10-2 m/ 2.83*105 m/s = 8.3*10-7 s
Since the initial speed is completely over the horziontal axis, the initial speed in the vertical axis is zero:
v2y = at :::::: a = 1.32*105 -j m/s / 8.3*10-7 s = 1.6*1011 -j m/s2
The force comes from the Newton's Second Law:
F = a*m = (1.5*10-26 kg)(1.6*1011 -j m/s2) = 2.4*10-15 -j N
So, the electric field would be:
E = F/q = 2.4*10-15 -j N/3.2*10-21 C = 750000 -j N/C
c. The force at the center of the loop should be only magnetic, since any electric field at the center of the loop would be zero, while the magnetic field should have a value of B = uI/2R (where u is the vacuum permeability).
Said magnetic field is over the axis of the loop. If we assume that: the speed vo is over the horizontal axis, the particle is going eastward, its charge is positive and the 'north' direction is towards the person who reads the page we know that the magnetic field should point upwards.
Therefore, the loop's normal vector was pointing upwards.