5

A 2.6 C charge moving in a magnetic field of strength 7.4 x10-3 T experiences a magnetic force of magnitude3.5 N. Calculate the speed of the charge.(Give your answe...

Question

A 2.6 C charge moving in a magnetic field of strength 7.4 x10-3 T experiences a magnetic force of magnitude3.5 N. Calculate the speed of the charge.(Give your answer in m/s but don't include the units.)

A 2.6 C charge moving in a magnetic field of strength 7.4 x 10-3 T experiences a magnetic force of magnitude 3.5 N. Calculate the speed of the charge. (Give your answer in m/s but don't include the units.)



Answers

A charge of $-8.3 \mu \mathrm{C}$ is traveling at a speed of $7.4 \times 10^{6} \mathrm{m} / \mathrm{s}$ in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is $52^{\circ} .$ A force of magnitude $5.4 \times 10^{-3} \mathrm{N}$ acts on the charge. What is the magnitude of the magnetic field?

Hello in this problem were given the amount of magnetic force, the magnitude of magnetic force that is acting on a charged particle as it moves perpendicular to a magnetic field. So if we look back at our equations, we know expression for the magnetic force that charged particle feels as it's moving through a magnetic field. It's the charged particle multiplied by its speed multiplied by the magnitude magnetic field, which is moving, multiplied by sine of the angle between the magnetic field and the velocity of the particle. But we were just told that that is 90 degrees because the particle is traveling perpendicular to the magnetic field. So in the end were given a value. I'm gonna label it as the magnetic force. One of I think it's 2.7 times 10 to the negative three Nunes, and that should be equal to the charge of my charge. Particle multiplied by its speed, then multiplied by the magnitude magnetic field because the sign of 90 degrees term goes to one. So we have this expression here. We know how all three of these properties of our situation the charge of the particle, its speed and the man into the field. All multiply to be a specific value, but were asked not about anything particular about the system rack. She asked about what the force was going to be on this charged particle. If, instead of it traveling perpendicular to a magnetic field, it travels at a new angle of 38 degrees. So all we have to do is just right. Our equation for that situation, right? We want to find the magnetic field magnetic force to That's acting on this particle when it's traveling at 38 degrees between right, the direction between the the angle between the direction of the motion of the charged particle and the magnetic field is now 30 degrees. So how will we write that out? Well, we would still need to know the charge of particle, but that really hasn't changed. We need another velocity of it. But we're told that that's not changed and the magnet to the magnetic field. And again, that hasn't changed. But now the angle between the magnetic field and the velocity is no longer 90 degrees. It's 38 degrees. But he just found that the combination of all three of these is given by this number 2.7 times 10 to the negative three units. So all we have to do to figure out this new force that the charged particle feels traveling at 38 degrees to the north magnetic field instead of 90 degrees in that field is our old force. The 2.7 times had negative free multiplied by sine of 38 degrees. We plug it into our calculator. We see that before the charged particle feels a force of 1.7 times 10 to the negative three Nunes, when it is traveling at 38 degrees to the magnetic field and set up 90 degrees, that is our answer.

So this question belongs to the magnetic forces in which we have magnetic field B equals to 0.45 Tesla. And for the part we have to calculate the magnitude and direction of the force on the charge Q equals to 0.10 micro column in the plus sign. Moving southward with the steel we equals to 3.5 m per second. So forth on the moving charge particle Q. It is equal to Q. V. B. Sign tita. Okay. So since Charles particle is moving in the north direction. Sorry, in the south direction. Okay. And the magnetic field be it is in the east directions of the angle tita becomes 90 degree. Which is in the which is the angle between velocity and magnetic field strength. So substituting values so we get magnitude of force F. This is equal to 0.10 more player bait. And to the power minus six school um speed we which is 3.5 m per second And B which is 0.45 and sign 90°.. So from here, magnitude of force F comes out to be 0.16 Molecular by 10. To the power minus six newton. Okay. And from the right hand, some rules, we get the direction which will be straight up straight up. Okay, so we can write this. This is the answer for the part of the problem magnitude as well as direction. Now, for the part B, we have to determine the magnitude of this force with charge Q equals two minus 0.10 micro column. So since here the nature of the charge has been changed. So the magnitude of the force will remain same. But this direction will get rivers. So we can right here the force F vector. This will be 0.16 more clear by 10. To the power minus six newton. In the straight down direction. Straight down direction. Okay, so this becomes the answer for the part B of the problem. Okay,

Belongs to the magnetic force in which we have a small particle carrying charge Q equals to 50 micro column. And it is moving with the speed we equals to 35 m per second. Through a uniform magnetic field be it is equals to 0.75 Tesla. The velocity and magnetic field both R. In the Xy plane. And with velocity making an angle 3- one that is equal to 25° with the. And magnetic field is making an angle Theta two equals 2 to 62 degree. Both measured counterclockwise from the X. Axis. Okay so the angle tita between magnetic field and velocity is equal to 2 62 degrees minus 25 degree. So this comes out to be 2 37 degrees. Okay To 37°.. So now we can write that the magnetic force on the charged particle, it is equal to Q. B. Signed tita. So substituting values so we will get that Q. Which is 15 player by 10 to the power minus six column is still we, which is 35 m per second magnetic field is G. 0.75 Tesla and sign to 37 degrees. So from here after solving we get fourth equals two minus 1.8 molecular weight and to the power minus three newton. And this force will be in the Y direction. Okay, so this is minus sign. So we can substitute, we can replace this minus and from here and we can apply this minus here. So we can write that the net force is in the uh minus Y direction. Okay, so this becomes the answer for this question. The direction has been obtained from the right and thumb rule. Okay, so this is the answer for this question. Okay.

For this problem. We're told that we have a charged particle with charge of 18 micro Coolum, and we're told that when it travels at 24 meters per second at a right angle to some magnetic field, the force is 2.8 times 10 to the negative for Newton's were asked to find that the force will be when that angle theta is no longer 90 degrees. But instead it's 25 degrees and the velocity is 6.3 meters per second. This debt. So what? We first start by doing well. We were looking for four. So let's write down the equation for force of a charged particle moving in a magnetic field. We know everything in this equation except for what the magnetic field is. So the first thing we have to do is we have to use the first information to calculate for the magnetic field. So we rearrange this equation, we get force divided by charge, speed and science data. In this case, which is a sign of 90 degrees, the force, we're told, is 2.8 times 10 to the negative for neatness just like that divided by the charge 18 Now it's micro Coolum. So you have to do 10 times 10 to the my native six Coolum. The speed is 24 minutes for a second. And if you work this out, what you get is you get around 0.65 Tesla for the magnetic field. Now that we know what the magnetic field strength is, we go force equal to our charge again is our 18 times 10 the negative six school. Um, our velocity in this case is 6.3 meters per second. Our magnetic field is there 0.65 Tessa and we've got sign of 25 degrees. And when you work that out, you find that our forces 3.1 times 10 to the negative five Newtons and that is the end of the quiet problem.


Similar Solved Questions

1 answers
Consider the differential equation y" ~xy' +y = 0.The indicial equation is r(r _ 1) = 0. The recurrence relation is Ck+1 (k+r+l)+(k+r)-C(ktr_1) = 0.A series solution corresponding to the indicial root r = 0 is Select the correct answer. Y1 -Zxo (~2x)* [1!(-1). 1.3..(2k-1)] b.J1 =*J-=Z;-o(-2x)* [k(2k-3)!] d:Y1 =xy=Z;-o(-2x)* [X1.3 ~(2k-3) ]
Consider the differential equation y" ~xy' +y = 0.The indicial equation is r(r _ 1) = 0. The recurrence relation is Ck+1 (k+r+l)+(k+r)-C(ktr_1) = 0.A series solution corresponding to the indicial root r = 0 is Select the correct answer. Y1 -Zxo (~2x)* [1!(-1). 1.3..(2k-1)] b.J1 =* J-=Z;-o...
5 answers
(be) +[~( HN)Sv] = (be) THN & + (be) 8V(s1d €) YIOM moK MO4S : [T(HNJSV] pue 'Bquoutue "UOL JAIIS JO SUOLIBIJUJJUO? UnuIqYInba a4} JuTua1ap [T(HNJBV] amnd jo uOqEIuaJUOJ W 00S"0 e YHIM SHEIS uognjos e JI 'L0*09 [ Jo an[eA 'Xe sey 4PIYM "UOIJEJI unpquba SuIMO[[of 341 01 SupJOJJE BJuOUue YJIM S1J831 UOF SV J4LSuaIO
(be) +[~( HN)Sv] = (be) THN & + (be) 8V (s1d €) YIOM moK MO4S : [T(HNJSV] pue 'Bquoutue "UOL JAIIS JO SUOLIBIJUJJUO? UnuIqYInba a4} JuTua1ap [T(HNJBV] amnd jo uOqEIuaJUOJ W 00S"0 e YHIM SHEIS uognjos e JI 'L0*09 [ Jo an[eA 'Xe sey 4PIYM "UOIJEJI unpquba SuIMO...
5 answers
82 + 128 + 1008 + 68F where F(8) 8+61/(s^2+842)82+128+100Therefore f(t)
82 + 128 + 100 8 + 6 8 F where F(8) 8+6 1/(s^2+842) 82+128+100 Therefore f(t)...
5 answers
'equilibrium constant for the reaction '#D The 2CrO42 2H+ Crz0,? - + Hzo 1044 The molar absorptivities for the two principal species in a solution is 4.2 X of K,Cr_Ov are
'equilibrium constant for the reaction '#D The 2CrO42 2H+ Crz0,? - + Hzo 1044 The molar absorptivities for the two principal species in a solution is 4.2 X of K,Cr_Ov are...
4 answers
Poxnts) Celeulate the sudactthe cylinder I? + y" 16 contained betwzen the planes
poxnts) Celeulate the sudact the cylinder I? + y" 16 contained betwzen the planes...
5 answers
Let z = f(x, y) and x = t-S,y = se'. (a) Find Zu: (b) If 2 = using Part (a) to find Zut: Vx +2y
Let z = f(x, y) and x = t-S,y = se'. (a) Find Zu: (b) If 2 = using Part (a) to find Zut: Vx +2y...
5 answers
07 Gu _ {By'0,13023I = (2y -19*/2,1 <y < 5What curve passes through_the point (1,5) and has arc length O the interval [2,6] given by J; V1 + T6r-odr?
07 Gu _ {By'0,13023 I = (2y -19*/2,1 <y < 5 What curve passes through_the point (1,5) and has arc length O the interval [2,6] given by J; V1 + T6r-odr?...
5 answers
Ashley has been training for a 10-kilometer race: Her average training pace is 8 minutes and 15 seconds per mile. If she maintains this pace during the race, which of the following will be her finishing time? (1 mile = 5280 feet; 1 foot = 0.3048 meters) A) 51.23 B) 50.53 C) 51.14 D) 82.30
Ashley has been training for a 10-kilometer race: Her average training pace is 8 minutes and 15 seconds per mile. If she maintains this pace during the race, which of the following will be her finishing time? (1 mile = 5280 feet; 1 foot = 0.3048 meters) A) 51.23 B) 50.53 C) 51.14 D) 82.30...
5 answers
An object executing SHM with an amplitude 14 cm. Find the magnitude of its displacement (in cm) when the value of its KE is equal to 1/3 th of its total energy: Write answer with 2 decimal places:Answer:
An object executing SHM with an amplitude 14 cm. Find the magnitude of its displacement (in cm) when the value of its KE is equal to 1/3 th of its total energy: Write answer with 2 decimal places: Answer:...
5 answers
10. (5) Provide an efficient synthetic rouc (not mechanism) for the following molecule, 2-oxopentanal , from methylcyclopentane_ Clearly show all reagents and intermediates.OHC
10. (5) Provide an efficient synthetic rouc (not mechanism) for the following molecule, 2-oxopentanal , from methylcyclopentane_ Clearly show all reagents and intermediates. OHC...
5 answers
Histogram 0f cholesterol level of 500 adults shown below:0.J5 1 0.25 L 0.45 0.05Chalaeterel Iaval Dtr AL HAItIILeulLutdt(1) Comment on the main feature Of the histogram (Mode; Center; Spread, Shape; Outller)(2) Wvhat percent of adults in the data have cholesterol level between 200 and 2052(3} How many adults the data have cholesterol levcl belaw 205?
histogram 0f cholesterol level of 500 adults shown below: 0.J5 1 0.25 L 0.45 0.05 Chalaeterel Iaval Dtr AL HAItIILeulLutdt (1) Comment on the main feature Of the histogram (Mode; Center; Spread, Shape; Outller) (2) Wvhat percent of adults in the data have cholesterol level between 200 and 2052 (3} H...
5 answers
Con-congexternal browser-O&launchUrl-https%2534252F*252FnewcSavedaltempts IeltCheck my workThe hydrolysis of ATP to ADP releases 7.3 kcalmol of energy: If the coupled reaction, fructose 6 phosphate ATP (ruetose 1,6-bisphosphate ADF; relenses 3.4 kcaV/mol of energy how much energy is required for the phosphorylation of fructose phosphate?Tructose 6-phosphate HPO;fructose [,6-bisphosphate Hzokcal/mol
con-congexternal browser-O&launchUrl-https%2534252F*252Fnewc Saved altempts Ielt Check my work The hydrolysis of ATP to ADP releases 7.3 kcalmol of energy: If the coupled reaction, fructose 6 phosphate ATP (ruetose 1,6-bisphosphate ADF; relenses 3.4 kcaV/mol of energy how much energy is required...
5 answers
23_ Use limit methods to determine which of the two given functions grOws faster, O state that they have comparable growth rates 11l0; eo.01x (b) 2?In ; In? ,Ii:
23_ Use limit methods to determine which of the two given functions grOws faster, O state that they have comparable growth rates 11l0; eo.01x (b) 2?In ; In? , Ii :...

-- 0.018541--