JEE MAIN - Physics (2004 - No. 34)
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
$${{{{\left( {B\pi r\omega } \right)}^2}} \over {2R}}$$
$${{{{\left( {B\pi {r^2}\omega } \right)}^2}} \over {8R}}$$
$${{B\pi {r^2}\omega } \over {2R}}$$
$${{{{\left( {B\pi r{\omega ^2}} \right)}^2}} \over {8R}}$$
Explanation
$$\phi = \overrightarrow B .\overrightarrow A ;\phi = BA\cos \,\omega t$$
$$\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$$
$$i = {{\omega BA} \over R}\sin \,\omega t$$
$${P_{inst}} = {i^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t$$
$${p_{avg}} = {{\int\limits_0^T {{P_{inst}} \times dt} } \over {\int\limits_0^T {dt} }}$$
$$ = {{{{\left( {\omega BA} \right)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}$$
$$ = {1 \over 2}{{{{\left( {\omega BA} \right)}^2}} \over R}$$
$$\therefore$$ $${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]$$
$$\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$$
$$i = {{\omega BA} \over R}\sin \,\omega t$$
$${P_{inst}} = {i^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t$$
$${p_{avg}} = {{\int\limits_0^T {{P_{inst}} \times dt} } \over {\int\limits_0^T {dt} }}$$
$$ = {{{{\left( {\omega BA} \right)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}$$
$$ = {1 \over 2}{{{{\left( {\omega BA} \right)}^2}} \over R}$$
$$\therefore$$ $${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]$$
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