Question

A rectangular loop of length l and breadth b is situated in a uniform magnetic field to induction B with its plane perpendicular to the field as shown in the figure.
Calculate the rate of change of magnetic flux and the induced e.m.f. if the loop is rotated with constant angular velocity $\omega ,$
(a) about an axis passing through the centre and perpendicular to the loop,
(b) about an axis passing through the centre parallel to the breadth through angle ${180}^o$.

Answer 1

(a) The flux linked through the loop at any instant is
$\phi =BA\cos \theta$
$=Blb\cos \theta =Blb=$ constant.
$\therefore$ Rate of change of magnetic flux is
$\frac{d\phi }{dt}=\frac{d}{dt}\left(Blb\right)\Rightarrow \frac{d\phi }{dt}=0$
Hence induced e.m.f. $\epsilon =-\frac{d\phi }{dt}=0$.
(b) when the loop is rotated through ${180}^o$ about an axis passing through centre and parallel to breadth then the change in magnetic flux
$d\varphi =-Blb-\left(Blb\right)=-2Blb$
$\therefore$ The induced e.m.f.
$e=-\frac{d\phi }{dt}\Rightarrow e=+\frac{2Blb}{\left(\pi /\omega \right)}$
$\Rightarrow e=\frac{2Blb\omega }{\pi }$.