Arectangular loop of wire with sides 0.240 and 0.554 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). the magnetic field has a magnitude of 0.579 t and is directed parallel to the normal of the loop's surface. in a time of 0.117 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. determine the magnitude of the average emf induced in the loop.

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amelvin41

20.02.2021

Physics

Given information is summarized as below:

$\omega _{o}=0.270rev/sec=1.69 rad/sec\\\alpha =0.895rev/sec=5.62rad/sec\\d=0.790m$

1)Using first equation of kinematics for angular motion we have

$\omega _{f}=\omega _{o}+\alpha t\\\\\therefore \omega _{f}=1.69+5.62\times 0.194\\\omega _{f}=2.780rad/sec$

2) Using second equation of kinematics for angular motion we have

$\theta =\omega _{o}t+\frac{1}{2}\alpha t^{2}\\\\\theta =1.69\times 0.194+05\times 5.62\times0.194^{2}\\\\\theta = 0.4336rad$

3) The tangential speed is given as

$v_{t}=\omega _{f}\times r\\\\v_{t}=2.78\times\frac{0.790}{2}=1.098m/s$

The resultant acceleration is given by

$a_{res}=\sqrt{(a_{tangential})^{2}+(a_{radial})^{2}}\\\\a_{res}=\sqrt{(\alpha r)^{2}+(\omega _{f}^{2}r)^{2}}\\\\a_{res}=3.77m/s^{2}$

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