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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jan 2007 16:06:28 IST
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A wire shaped as a semicircle of radius 'a' rotates about an axis OO' with an angular velocity  in a uniform magnetic field of induction B. The rotation axis is perpendicular to the field direction. The total resistance of the circuit is equal to R. Neglecting the magnetic field of induced current, find the mean amount of thermal power being generated in the loop during a rotation period.
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21 Jan 2007 19:08:25 IST
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mean thermal power= (1/2) E^2/R
emf induced across the 2 ends of the semicircle = 2a^2BW
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21 Jan 2007 21:43:14 IST
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i dont think thats correct
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23 Jan 2007 16:18:02 IST
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B2a4w2/4R
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24 Jan 2007 19:55:56 IST
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The induced emf is given by
e = -d(BA)/dt = -B dA/dt
Since angular velocity is w, so we can find rate of change of area i.e dA/dt
So here e = (1/2)B w a2
Here resistance of the wire = R
so induced current = i = e/R = (B w a2 / 2R)
Thus thermal power generated in the loop = i2R,
substitute the value of i to obtain the desired result.
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2 Jun 2008 13:58:29 IST
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The intersting thing is that there is NO INDUCED EMF & HENCE CURRENT !!!
It is because there is no change in flux wrt time as the semicircular wire rotates in a UNIFORM MAGNETIC FIELD !!
hence mean amount of thermal power generated = 0 watts
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