Consider a conducting rod of length 34 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane
1 answer:
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Answer:
This is your answer
Explanation:
Actually I took this from go ogle
Answer:
a) Final position is x = 0.90 m
b) Final position is x = 0.133 m
Explanation:
The workdone between two points is usually approximated as the area under the force-distance curve between those two points.
From the graph,
As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,
The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J
The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m
Work done under this bar = 0.4 × 0.25 = 0.1 J
Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J
So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N
0.21 = 0.18 + 0.2 (x - 0.75)
0.03 = 0.2x - 0.15
0.2x = 0.18
x = 0.9 m
Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.
b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.
From the starting point where the initial position is 0.40 m, the force here is 0.80 N
The workdone under this bar to the left is
The workdone = 0.8 (0.25 - 0.4) = - 0.12 J
Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N
- 0.19 = -0.12 + 0.6 (x - 0.25)
-0.07 = 0.6x - 0.15
0.6x = 0.08
x = (0.08/0.6) = 0.133 m
Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.
Hope this Helps!!!
