An annular circular brass disk of inner radius r and outer radius R is rotating about an axis passing through its center and perpendicular to its plane with a uniform angular velocity - in a uniform magnetic field of induction B normal to the plane of the disk- The induced emf between the inner and the outer edge of the annular disk is -

The correct option is B Bω (R^2 r^2)2 d( emf ) = Bwxdx emf= ∫ _ r ^ R Bwxdx = Bw ∫_r^R x ^ 2 2 =Bw( R ^ 2 r ^ 2 2 ) Emf between ...

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Standard XII

Physics

Lenz's Law

An annular ci...

Question

An annular circular brass disk of inner radius $r$ and outer radius $R$ is rotating about an axis passing through its center and perpendicular to its plane with a uniform angular velocity $ω$ in a uniform magnetic field of induction $B$ normal to the plane of the disk. The induced emf between the inner and the outer edge of the annular disk is :

\frac{B ω (r^{2} + R^{2})}{2}

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\frac{B ω (R^{2} - r^{2})}{2}

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\frac{B ω (r - R)}{2}

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\frac{B ω (r + R)}{2}

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Solution

The correct option is B $\frac{B ω (R^{2} - r^{2})}{2}$
$d (e m f) = - B w x d x$
$e m f = - \int_{r}^{R} B w x d x$
$= - B w \int_{r}^{R} \frac{x^{2}}{2}$
$= B w (\frac{R^{2} - r^{2}}{2})$
Emf between inner and outer edge will be
$= B w (\frac{R^{2} - r^{2}}{2})$

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The correct option is B Bω(R2−r2)2d(emf)=−Bwxdxemf=−∫RrBwxdx=−Bw∫Rrx22=Bw(R2−r22)Emf between inner and outer edge will be=Bw(R2−r22)

The correct option is B $\frac{B ω (R^{2} - r^{2})}{2}$
$d (e m f) = - B w x d x$
$e m f = - \int_{r}^{R} B w x d x$
$= - B w \int_{r}^{R} \frac{x^{2}}{2}$
$= B w (\frac{R^{2} - r^{2}}{2})$
Emf between inner and outer edge will be
$= B w (\frac{R^{2} - r^{2}}{2})$