Question 129: Electromagnetic Induction

Question

In a uniform magnetic field of induction $B$ a wire in the form of a semicircle of radius $r$ rotates about the diameter of the circle with an angular frequency $\omega .$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R,$ the mean power generated per period of rotation is

Accepted Answer

$\phi  = \overrightarrow B .\overrightarrow A ;\phi  = BA\cos \,\omega t$
$\varepsilon  =  - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$
$i = {{\omega BA} \over R}\sin \,\omega t$
${P_{inst}} = {i^2}R = {\left( {{{\omega BA} \over R}} ight)^2} 	imes R{\sin ^2}\omega t$
${p_{avg}} = {{\int\limits_0^T {{P_{inst}} 	imes dt} } \over {\int\limits_0^T {dt} }}$
$ = {{{{\left( {\omega BA} ight)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}$
$ = {1 \over 2}{{{{\left( {\omega BA} ight)}^2}} \over R}$
$	herefore$ ${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} ight)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} ight]$

Answer

${{{{\left( {B\pi r\omega } \right)}^2}} \over {2R}}$

Answer

${{{{\left( {B\pi {r^2}\omega } \right)}^2}} \over {8R}}$

Answer

${{B\pi {r^2}\omega } \over {2R}}$

Answer

${{{{\left( {B\pi r{\omega ^2}} \right)}^2}} \over {8R}}$

Electromagnetic Induction

Choose the correct answer: