Asked by Mary
Interactive LearningWare 22.2 at wiley/college/cutnell reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m 0.55 m. The magnetic field has a magnitude of 2.1 T and is inclined at an angle of 75° with respect to the normal to the plane of the loop.
(a) If the magnetic field decreases to zero in a time of 0.46 s, what is the magnitude of the average emf induced in the loop?
V
(b) If the magnetic field remains constant at its initial value of 2.1 T, what is the magnitude of the rate A / t at which the area should change so that the average emf has the same magnitude as in part (a)?
m2/s
Please tell me where I went wrong.
This is what I did but it is incorrect.
a.
initial Phi= 75 degrees
= (2.1T)(0.1925 m^2)(cos 75 degrees)
= 0.104627
Final Phi= 0 degrees
= (2.1T)(0.1925 m^2)(cos 0 degrees)
= 0.40425
delta Phi = 0.104627 - 0.40425
= -0.299622401
E= -N (delta Phi/delta t)
E= -1(-0.299622401/0.46 s)
E= 0.651353046 V
Can you please explain part b.
(a) If the magnetic field decreases to zero in a time of 0.46 s, what is the magnitude of the average emf induced in the loop?
V
(b) If the magnetic field remains constant at its initial value of 2.1 T, what is the magnitude of the rate A / t at which the area should change so that the average emf has the same magnitude as in part (a)?
m2/s
Please tell me where I went wrong.
This is what I did but it is incorrect.
a.
initial Phi= 75 degrees
= (2.1T)(0.1925 m^2)(cos 75 degrees)
= 0.104627
Final Phi= 0 degrees
= (2.1T)(0.1925 m^2)(cos 0 degrees)
= 0.40425
delta Phi = 0.104627 - 0.40425
= -0.299622401
E= -N (delta Phi/delta t)
E= -1(-0.299622401/0.46 s)
E= 0.651353046 V
Can you please explain part b.
Answers
Answered by
bobpursley
I am wondering what äverage'means here. If it is really average, then you would have to integrate over time, and divide by the time.
Answered by
Mary
Please clarify.
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