I keep getting this wrong? What is the right combination of answers? Course Cont
ID: 1778343 • Letter: I
Question
I keep getting this wrong? What is the right combination of answers?
Course Contents» » Set 6 (due Wed 10/18 at 11:59PM)Loop of Wire in a Field O Timer Notes Evaluate Feedback Print A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B field (B is constant in time) and then back into a field free region to the left. The self inductance of the loop is negligible True When entering the field the coil experiences a magnetic force to the right. FalseWhen leaving the field the coil experiences a magnetic force to the left True Upon leaving the field, a counterclockwise current flows in the loop FalseUpon entering the field, a counterclockwise current flows in the loop SubmitAnswer You have entered that answer before Incorrect. Tries 6/8 Previous TriesExplanation / Answer
According to the given problem,
A)When entering the field the coil experiences a magnetic force to the right.
Ans: True, when the loop is moving into of the field, the left hand side will be out of
the magnetic field so no force will be on it. The current will be moving counterclockwise
to counter the decrease in flux. The force on the top and bottom portions of the
loop will cancel. The only remaining part is the right hand side. F = IL × B
gives the force as pointing to the left. This can also be deduced by realizing
that you must do work to pull the loop out of the field. Thus, a force must be
pulling the loop out of the field.
B)When leaving the field the coil experiences a magnetic force to the left.
Ans: False, when the loop is moving out of the field, the left hand side will be out of
the magnetic field so no force will be on it. The current will be moving clockwise
to counter the decrease in flux. The force on the top and bottom portions of the
loop will cancel. The only remaining part is the right hand side. F = IL × B
gives the force as pointing to the right. This can also be deduced by realizing
that you must do work to pull the loop out of the field. Thus, a force must be
pulling the loop into the field.
C)Upon leaving the field, a counterclockwise current flows in the loop.
Ans: False, see (B)
D)Upon entering the field, a counterclockwise current flows in the loop.
Ans: True,The current will travel counterclockwise to counter the increase in flux. The only portion of the loop in the field that matters is the left hand side which will produce a force to the right.