Induced EMF: Find Flux for Rectangular Loop in z-y Plane

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The discussion centers on calculating the induced electromotive force (emf) in a rectangular loop situated in the z-y plane, with one rod moving in the x-direction at a specified velocity. The magnetic field is defined as B = (10/x) cos(100t) j, and the flux through the loop is expressed using a double integral. A correction is noted regarding the limits of integration, specifying that the lower limit for x should be 1, not 0, while the upper limit is adjusted to account for the rod's movement. Participants confirm that the equation can be simplified and integrated to find the induced emf. The focus remains on ensuring the accuracy of the flux equation for further calculations.
robert25pl
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Can someone check my equation for flux. Thanks

A rectangular loop in the z-y plane is situated at t = 0 at the points (x=1,z=0), (x=1,z=5), (x=4,z=5), and (x=4,z=0). The rod of the loop with end points (x=4,z=0), and (x=4,z=5) is moving in the x-direction with a velocity
v=5\vec{i}m/s while the rest of the loop remains fixed. Find induced emf in the loop for all t

B = (10/x) cos100t j

\psi=\int_{s}B\cdot\,ds=\int_{xo=1}^{xo=4} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}
 
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robert25pl said:
Can someone check my equation for flux. Thanks

A rectangular loop in the z-y plane is situated at t = 0 at the points (x=1,z=0), (x=1,z=5), (x=4,z=5), and (x=4,z=0). The rod of the loop with end points (x=4,z=0), and (x=4,z=5) is moving in the x-direction with a velocity
v=5\vec{i}m/s while the rest of the loop remains fixed. Find induced emf in the loop for all t

B = (10/x) cos100t j

\psi=\int_{s}B\cdot\,ds=\int_{x=0}^{xo+5t} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}

Looks OK to me except the lower limit on the x integral should be 1, not zero, with xo = 4 in the upper limit. I assume you can simplify and integrate this.
 
Yes I can do that and then find emf.
 
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\psi=\int_{s}B\cdot\,ds=\int_{x=1}^{4+5t} \int_{z=0}^{5}\frac{10}{x}cos100t\vec{j}\cdot\, dx\,dz\vec{j}
 
Last edited:
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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