Question 146: 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

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