At a point on the Earth’s surface, the Earth’s magnetic field points exactly North with magnitude 33.7μT.

Sunday, 26 August 2018

1:19 AM

Untitled picture.png Machine generated alternative text:
Question:
part 1)
-19 c
At a point on the Earth's surface, the Earth's magnetic field points exactly North with magnitude 33.7"T. The charge of a deuteron (a panicle comprising a proton plus a neutron) is 1.60 x 10
and its mass is 3.34 x 10-27 kg. What is the minimum speed v at which a deuteron must travel so that the magnetic torce on it and its weight add to zero?
part 2)
What is the direction of its velocity for the magnetic force and the weight to cancel witn that minimum panicle speed v?
(No answer given)
part 3)
An electron has velocity i,' =
Check
x 105i + 3.0 x 1059 ms—I. A magnetic field B = 2.3i + 1.8Q T. Give an expression tor the force F on the electron due to the magnetic field.

Untitled picture.png Machine generated alternative text:
Feedback tor Part 1)
use the definition of magnetic force:
so, F = qvBsin(9).
For mechanical equilibrium, mg = quBsin(9).
For minimum value:
qöxÉ
qB sin(9) •
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11
Feedback tor Part 2)
B is North. F is up. F = x B , so must be East. (It the thumb ot 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 b.)
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11
Feedback tor Part 3)
You have correctly entered the first component
You have incorrectly entered the second component
You have correctly entered the third component.
+ x (Bri + BS and (axi + a.vj + azü) x (bri + buj + = (aubz — azbu)f + (azbx
— axbz)j + (azbu — avbz)ü.
-1.2 x 10
-145 + N.
Marks for this submissiom 0.22/0.33. This submission attracted a penalty of 0.11

Untitled picture.png Machine generated alternative text:
Feedback tor Part 1)
use the definition of magnetic force:
so, F = qvBsin(9).
For mechanical equilibrium, mg = quBsin(9).
For minimum value:
qöxÉ
qB sin(9) •
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11
Feedback tor Part 2)
B is North. F is up. F = x B , so must be East. (It the thumb ot 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 b.)
Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11
Feedback tor Part 3)
You have correctly entered the first component
You have incorrectly entered the second component
You have correctly entered the third component.
+ x (Bri + BS and (axi + a.vj + azü) x (bri + buj + = (aubz — azbu)f + (azbx
— axbz)j + (azbu — avbz)ü.
-1.2 x 10
-145 + N.
Marks for this submissiom 0.22/0.33. This submission attracted a penalty of 0.11

Untitled picture.png Machine generated alternative text:
A correct answer is 0.0061, which can be typed in as follows:
e. oe61
A correct answer
'East"
A correct answer is 0, which can be typed in as follows:
A correct answer —1.2e — 14, which can be typed in as follows:
-1.22-14
A correct answer is 0, which can be typed in as follows:

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