Question

A semicircle loop $PQ$ of radius ${}^{\prime }{R}^{\prime }$ is moved with velocity ${}^{\prime }{v}^{\prime }$ in transverse magnetic field as shown in figure. The value of induced emf. at the end of loop is :-

Answer 1

Induced $\begin{array}{c}emf=\frac{-d\theta }{dt}\\ E=\frac{-d}{dt}\left(BA\cos \theta \right)\\ E=-B\frac{dA}{dt}\cos \theta \end{array}$ because the ring falls with its vertical plane in horizontal mehnetic induction $\begin{array}{c}\overrightarrow{B}.\theta =0and\cos \theta =1\\ orE=-B.\frac{d}{dt}\left(2Rdx\right)\\ E=-2RB\left(\frac{dx}{dt}\right)\\ \left|E\right|=2RBV\end{array}$

This is equal to the emf developed in an imaginary rod joining M &Q

According to fleming right hand rule, M is at low potential and Q is at high potential and induced emf is from M to Q.