Solving for Magnetic Flux in a Rotating Wire Loop

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The discussion focuses on calculating the change in magnetic flux for a wire loop with a semicircular section when it is rotated through half a revolution in a constant magnetic field of 0.75 T. The initial calculation of the change in flux is derived using the formula Change in flux = BA[cos(final angle) - cos(initial angle)], resulting in a value of -0.094 Tm². A key point of confusion is the interpretation of the final angle being 180 degrees, as the user struggles to understand why the angle is not zero despite the loop remaining parallel to the screen. It is clarified that the change in flux is attributed to the change in the area of the loop rather than the angle itself, as the semicircle's orientation flips during the rotation. The discussion emphasizes the importance of understanding the area change in magnetic flux calculations.
lovelyrwwr
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1. A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius . The normal to the plane of the loop is parallel to a constant magnetic field of magnitude 0.75 T. What is the change in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?



Homework Equations


A = Pi(radius^2) / 2 = (Pi)(0.2^2)/2 = 0.0628
Flux = BA

The Attempt at a Solution


Change in flux = final flux - original flux = BA[cos(final angle) - cos(initial angle)]
Change in flux = 0.75(0.0628)[cos180-cos0] = -0.094 Tm^2


I already know that the answer is -0.093 Tm^2.

But I am unsure how why the final angle is 180 such that you get a flux that is negative as calculated below. I guess I just cannot conceptualize WHY the the angle is 180 degrees when the wire goes through half of a revolution. I mean, when it goes through this half-revolution, the plane is still parallel to the screen of the computer. Thus, isn't the angle between plane of the computer screen and the magnetic field (which goes into the screen of the computer) still 0?
 
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Yes here it is thank you!
 
lovelyrwwr said:
But I am unsure how why the final angle is 180 such that you get a flux that is negative as calculated below. I guess I just cannot conceptualize WHY the the angle is 180 degrees when the wire goes through half of a revolution. I mean, when it goes through this half-revolution, the plane is still parallel to the screen of the computer. Thus, isn't the angle between plane of the computer screen and the magnetic field (which goes into the screen of the computer) still 0?
The change in flux is due to the change in the area of the loop, not in any change in angle. Originally, the area of the loop is a rectangle plus a semicircle. But when the semicircle flips over, the area is now the rectangle minus a semicircle.
 
Wow - I pay for chegg to understand solutions to problems. It misled me to believe that it had to do with angle.

Thank you so much Doc Al. You always pull through!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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