Aconducting loop with area 0.15 m2 and resistance 6.0 ω lies in the xx-yy plane. a spatially uniform magnetic field points in the zz-direction. the field varies with time according to = − where = 2.0 t/s2 and = 8.0 t. find the loop current (a) at = 3.0 s and (b) when = 0.
a.) is the correct answer.
answer: good job it's all correct !
the answer for apex is the lowest
the correct answers are:
ab ⇒ not moving
bc ⇒ positive acceleration
cd ⇒ constant velocity
de ⇒ not moving
ef ⇒ negative acceleration
assumption: let me assume that the horizontal axis represents the time (as the image is cropped from below). [i know, although, that it is time in x-axis] in that case, it is a position vs. time graph.
first, as you can see in the graph that from point a to point b, the position of the body is same. to be more precise, as you can see in the interval 0-20, the position is 20 meters at every point, which means that the body is not moving. hence, the path ab corresponds to not moving.
second, from point b to c, as you can see that the position of a body from intervel 20 to (about) 35, the position of a body has started changing slowly from the "not moving" state; however, between 35 and 40, the position changes rapidly, which means that the velocity of the body is increasing. if the velocity of the body increases, there will be a positive acceleration. hence, the path bc corresponds to positive acceleration.
third, from point c to d, the position of a body is changing constantly. it means that there will be a constant velocity from point c to point d. hence, the path cd corresponds to constant velocity.
fourth, from point d to e, the position is again not changing. it means that the body is in not moving state. hence, the path de corresponds to not moving.
last, from point e to f, as you can see, the body's position is changing rapidly at start, but after about 85 (seconds), the position is changing slowly, which means that the velocity of the body is decreasing. if the velocity of the body is decreasing, there will be a negative acceleration. hence, the path ef corresponds to negative acceleration.