A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electrom
1 answer:
Answer:
a) F = 0.01234N to the south direction
b) F = 0.01234N to the north direction
c) F = 0.01234N
d) North of West
Explanation:
The magnetic force in a magnetic field is given by:
Length of the wire, l = 1 cm = 0.01 m
Current flowing through the wire, I = 2.15 A
a) If the magnetic field direction is east
∅ = 90
F = 0.574 * 2.15 * 0.01
F = 0.01234 N to the south direction according to the Fleming's Right Hand Rule
b) If the magnetic field direction is south
The magnitude of the magnetic force remains the same
That is , F = BIL sin 90
F = 0.01234 N to the west
c) If the magnetic field direction is 30 degrees south
The angle between the magnetic field and the length of the wire still remains 90 degrees
Therefore the magnitude of the force still remains 0.01234
F = 0.01234 N
d) the direction of the force is the North of West
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Explanation:
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that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 () and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get , we do:
And we substitute the values: