MCQ
A rectangular conducting loop of sides $8\, cm$ and $2\, cm$ with a small cut is moving out a region of uniform magnetic field of magnitude $0.3\, T$ directed normal to the loop as shown in fig $(i)$ and $(ii)$. If the velocity of loop is $1\,cm\, s^{-1}$, then the ratio of voltage developed across ab in case $(i)$ to case $(ii)$ is......$V$
  • A
    $2$
  • B
    $0.25$
  • $4$
  • D
    $1$

Answer

Correct option: C.
$4$
c
In case $(i)$ $V_{1}=B \ell v \quad(\ell=8 \mathrm{\,cm})$

In case $(iii)$ $\mathrm{V}_{2}=\mathrm{B}$ $. bv$ $\quad(\mathrm{b}=2 \mathrm{\,cm})$

$\therefore \frac{\mathrm{V}_{1}}{\mathrm{V}_{2}}=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{Bbv}}=\frac{\ell}{\mathrm{b}}=\frac{8}{2}=4$

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