MCQ
A rectangular loop has a sliding connector $PQ$ of length $2\ m$ and resistance $10\ \Omega  $  and it is moving with a speed $5\ m/s$ as shown. The set-up is placed in a uniform magnetic field $3\ T$ going into the plane of the paper. The three currents $I_1$ , $I_2$ and $I$ are 
  • A
    $I_1$ = $I_2$ = $3\ A$ , $I$ = $1\ A$
  • B
    $I_1$ = $I_2$ = $5\ A$ , $I$ = $2\ A$
  • $I_1$ = $I_2$ = $1\ A$ , $I$ = $2\ A$
  • D
    $I_1$ = $I_2$ = $I$ = $2\ A$

Answer

Correct option: C.
$I_1$ = $I_2$ = $1\ A$ , $I$ = $2\ A$
c
$I = \frac{{BVl}}{{15}}$

${I_1} = {I_2} = \frac{I}{2}$

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