6th December 2008

JEE Mains 2015 Guidance › Question Of the Day questions › 6th December 2008

Near Mid-Night question of the day [6]

A wire shpaed as a semi circle of radius a rotates about axis OO' with an angular velocity Ï‰ 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 the induced current, find the mean amount of thermal power being generated in the loop during a rotation period.

Image in the following post...

**** Edit****
Another one Shamelessly lifted from IE Irodov ;)
*** End Edit ***

#Miscellaneous #Question Of the Day

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8 Answers

62

Lokesh Verma ·2008-12-06 09:09:39

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1

satan92 ·2008-12-06 09:17:47

in time t the angle between the B and area= Î¸
Einduced= -d(Ï†b)/dt=-Bd(Acoswt)/dt=BAwsinwt=E₀sinwt

so thermal power= ₀âˆ«^2Ï€/wE₀sinwt*idt
but i= E₀sinwt/R

so total thermal power(after calculating )= (E₀²/2R )T

hence mean thermal power = E₀²/2R

sry forgot to mention E₀=BÏ€a²w/2

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62

Lokesh Verma ·2008-12-06 09:38:41

you have made a very small mistake.. in just some constants!!!

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1

satan92 ·2008-12-06 09:44:46

not getting that small one!!

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satan92 ·2008-12-06 09:52:46

please refer to it
not getting the mistake ..

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1

Venkatesh Adithya ·2008-12-06 21:05:43

hey area is A/2 since its a semi circle so that's it...

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satan92 ·2008-12-07 00:39:31

but in my answer A = Ï€a²/2 !!

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Celestine preetham ·2009-04-29 07:16:23

me too getting same ans as SATAN wats the mistake ?????

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Your Answer

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