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  • I have a doubt, kindly clarify. - Electromagnetic Induction and Alternating currents - JEE Main

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 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

  • Option 1)

    \frac{B\pi r^{2}\omega }{2R}\;

  • Option 2)

    \; \; \frac{(B\pi r^{2}\omega)}{8R}^{2}\; \;

  • Option 3)

    \; \frac{(B\pi r\omega)}{2R}^{2}\; \;

  • Option 4)

    \; \frac{(B\pi r\omega^{2})}{8R}^{2}

 

Answers (1)

best_answer

As we learnt in

Motional E.m.f due to rotational motion -

Semicircle conducting loop

\varepsilon = \frac{1}{2}Bwr^{2}

i = \frac{Bwr^{2}}{2R}

 

 

=\frac{(B\pi r^{2}\omega )^{2}}{4R}\times \frac{1}{2}=\frac{(B\pi r^{2}\omega )^{2}}{8R}

At any instant magnetic flux associated with the coil is

\phi =B.\frac{\pi r^{2}}{2}coswt

induced current =\frac{\varepsilon }{R}

= - \frac{d\phi }{dt}.\frac{1}{R}=\frac{B\pi\:r^{2}\omega sin\omega t}{2R}

Instantaneous power = I2R

=\left ( \frac{B^{2}\pi^2\:r^{4}\omega^2 sin^2\omega t}{4R^2}\right )R

Average power

= \frac{B^{2}\pi^2\:r^{4}\omega^2} {4R}<sin^2\omega t>

= \frac{B^{2}\pi^2\:r^{4}\omega^2} {8R}

\because <sin^2\omega t>=\frac{1}{2}

 


Option 1)

\frac{B\pi r^{2}\omega }{2R}\;

This option is incorrect

Option 2)

\; \; \frac{(B\pi r^{2}\omega)}{8R}^{2}\; \;

This option is correct

Option 3)

\; \frac{(B\pi r\omega)}{2R}^{2}\; \;

This option is incorrect

Option 4)

\; \frac{(B\pi r\omega^{2})}{8R}^{2}

This option is incorrect

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Plabita

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