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 filed of induction B normal to the plane of the disk. The induced emf between the inner and outer edge of the annular disk is

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 filed of induction B normal to the plane of the disk. The induced emf between the inner and outer edge of the annular disk is (A) (Bω(r2+R2)/2) (B) (Bω(R2-r2)/2) (C) (Bω(r-R)/2) (D) (Bω(r+R)/2)

Tardigrade Admin · Accepted Answer

d ( emf )=- Bwxdx emf =-∫ limits r R Bwxdx =- Bw ∫ limits r R ( x 2/2) = Bw (( R 2- r 2/2)) Emf between inner and outer edge will be = Bw (( R 2- r 2/2))

$\frac{B ω ( r ^{2} + R ^{2} )}{2}$

$\frac{B ω ( R ^{2} - r ^{2} )}{2}$

$\frac{B ω ( r - R )}{2}$

$\frac{B ω ( r + R )}{2}$

$d (e m f) = - Bw x d x$
$e m f = - r \int R Bw x d x$
$= - Bw r \int R \frac{x ^{2}}{2}$
$= Bw (\frac{R ^{2} - r ^{2}}{2})$
Emf between inner and outer edge will be
$= Bw (\frac{R ^{2} - r ^{2}}{2})$

2Bω(r2+R2)​

2Bω(R2−r2)​

2Bω(r−R)​

2Bω(r+R)​

d(emf)=−Bwxdx emf=−r∫R​Bwxdx =−Bwr∫R​2x2​ =Bw(2R2−r2​) Emf between inner and outer edge will be =Bw(2R2−r2​)

$\frac{B ω ( r ^{2} + R ^{2} )}{2}$

$\frac{B ω ( R ^{2} - r ^{2} )}{2}$

$\frac{B ω ( r - R )}{2}$

$\frac{B ω ( r + R )}{2}$

$d (e m f) = - Bw x d x$
$e m f = - r \int R Bw x d x$
$= - Bw r \int R \frac{x ^{2}}{2}$
$= Bw (\frac{R ^{2} - r ^{2}}{2})$
Emf between inner and outer edge will be
$= Bw (\frac{R ^{2} - r ^{2}}{2})$