4. Refer to the diagram here (not to scale)—and note the coordinate axes. Note a

Question

4. Refer to the diagram here (not to scale)—and note the coordinate axes. Note also: ?0° is eastward and ?90° is upward.

A particle of known mass m, initially moving eastward at an unknown speed v0, 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 B1. Then it enters the known uniform magnetic field B2 at a known velocity v2. As it enters B2, a known magnetic force F is exerted on the particle.

The data: m = 1.50 x 10-26 kg          L = 23.4 cm          B1 = 56.7 µT          B2 = 89.0 µT?150°      v2 = 3.12 x 105 m/s?25.0°          F = 7.28 x 10–18 N north          (All gravitational forces are negligible.)

a. Describe the particle’s travel within B1: 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 B1.

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 v0, 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 x 10–18 N north), in what direction was the loop’s normal vector pointing?

4. Refer to the diagram here (not to scale)-and note the coordinate axes Note also: L0° is eastward and L90s upward up east A particle of known mass m, initially moving eastward at an unknown speed v,, 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 m = 1.50 x 1026 kg = 3.12 x 105 m/s B. = 567pT 7.28 x 10-18 N north The data: 23.4 cm B. = 89.0 TL150 (All gravitational forces are negligible.) 25.0 F 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? alculate the total time the particle spent within b. 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 vo, 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 x 10-18 N north), in what direction was the loop's normal vector pointing? c.

Explanation / Answer

Hi,

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*105i 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.

I hope it helps.

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