MCQ
A metal wire $PQ$ slides on parallel metallic rails having separation $0.25\ m$ , each having negligible resistance .There is a $2\,\Omega $ resistor and $10\ V$ battery as shown in figure . There is a uniform magnetic field directed into the plane of the paper of magnitude $0.5\ T$ . A force of $0.5\ N$ to the left is required to keep the wire $PQ$ moving with constant speed to the right. With what speed is the wire $PQ$ moving ?........$m/s$ (Neglect self inductance of the loop)
  • A
    $8$
  • $16$
  • C
    $24$
  • D
    $32$

Answer

Correct option: B.
$16$
b
Induced $\mathrm{e}.$ $\mathrm{m.} \mathrm{f}.$ $=\mathrm{B} \ell \mathrm{v}=0.125 \mathrm{\,V}$

current $1=\frac{10-\mathrm{e}}{\mathrm{R}}=\frac{10-0.125 \mathrm{\,V}}{2}$

Force $B1\ell $

$=0.5\left(\frac{10-0.125 \mathrm{V}}{2}\right) 0.25=0.5 \mathrm{\,N}$ (given)

Solving $\mathrm{V}=16 \mathrm{\,m} / \mathrm{\,s}$

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