A proton (mass 1.67×10−27 kg) is travelling with v =1.88×106i^ ms−1 in an electric field E =3.32×105j^ Vm−1

Sunday, 26 August 2018

12:50 AM

Machine generated alternative text:
Question: 
part 1) 
A proton (mass 1.67 x 10 27 kg) is travelling witn = 1.88 x 106i ms¯l in an electric field E 
travelling in a straight line. Neglect all other forces in comparison witn electric and magnetic. 
part 2) 
What is the direction of that magnetic field? 
(No answer given) 
part 3) 
= 3.32 x vm 
What is the smallest magnitude B of magnetic tield required to keep it 
A proton is travelling at constant speed in a circle in the x, y plane witn radius r = 13.3 mm. It completes one circle in a time 1.74" s. The only non-negligible torce acting on it is due to a uniform 
magnetic field B. Give an expression tor the magnitude ot Bz, the z component of the magnetic field acting on it. 
Check

 

Machine generated alternative text:
Feedback tor Part 1) 
Fmagnetic = qvB sin(9) so: 
So the smallest B is in the case where sin(9) 
qvsin(B) • 
1. Here, for straight line motion, Newton's second law requires Felectric — 
magnetic, so their magnitudes are equal, so 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 
Feedback tor Part 2) 
qi x B must balance qE, where is in the direction and E is in the y directiom k x = j so x k — —j , so positive in the direction and positive B in the z direction give a magnetic 
force on a positive charge that is in the negative y direction and so can balance Eq. B in the positive z direction. (Itthe tnumö of your right hand is in the direction of and your right index finger in 
the b direction, then the middle finger of that hand, it at right angles to the others, IS in the direction otä x 
Figure 1 _ The electric field E is in the -Fy direction and particle travels in the -ex direction witn velocity v. 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11

 

Machine generated alternative text:
Feedback tor Part 3) 
The proton is travelling in a circle at uniform speed so the total force on it is centripetal witn magnitude FC = m 
This torce is always in the x, y plane, so the magnetic field must be at right angles to that plane,i.e_ in the z direction. The centripetal force is at right angles to F, so ts magnitude is 
qvB sin(9) = qvBz, so: 
qt'Bz = m 
The time taken for one circle is T 
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 
A correct answer is 0.177, which can be typed in as follows: 
e. 177 
A correct answer 
"k" 
A correct answer is 0.0378, which can be typed in as follows: 
e. 8378 
27r m

 

 

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