Question

A metal disc of radius ‘R’ rotates with an angular velocity ‘ω’ about an axis perpendicular to its plane passing through its centre in a magnetic field of induction ‘B’ acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only)

• A. $\frac {BωR}{2}$
• B. $\frac{Bω^2R^2}{2}$
• C. $\frac {BωR ^2}{2}$
• D. $\frac {Bω^2R}{2}$

Answer 1

Answer: (C)

$\frac {BωR ^2}{2}$

Answer 2

Area A=πR2
Now differentiate the area
$\frac {dA}{dt}$ = $\frac {πR^2}{2π/ω}$
$\frac {dA}{dt}$t = $\frac {ωR^2}{2}$
e.m.f = - B x $\frac {dA}{dt}$
e.m.f = $\frac {BωR^2}{2}$
So, the correct option is (C) $\frac {BωR^2}{2}$