Q152 / 160 · AP EAMCET 2024 (22 May Shift 2) · Physics

A metallic wire loop of side $(l) 0.1 \mathrm{~m}$ and resistance of $1 \Omega$ is moved with a constant velocity in a uniform magnetic field of $2 \mathrm{Wm}^{-2}$ as shown in the figure. The magnetic field is perpendicular to the plane of the loop. The loop is connected to a network of resistors. The velocity of loop so as to have a steady current of 1 mA in loop is Figure

  • A
    $0.67 \mathrm{~cm} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $2 \mathrm{~cm} \mathrm{~s}^{-1}$
  • C
    $3 \mathrm{~cm} \mathrm{~s}^{-1}$
  • D
    $4 \mathrm{~cm} \mathrm{~s}^{-1}$

Solution

$1=0.1 \mathrm{~m}, \mathrm{R}=1 \Omega, \mathrm{~B}=2 \mathrm{Wm}^{-2}, \mathrm{I}=1 \mathrm{~mA}=10^{-3} \mathrm{~A}$ Motional emf, $\mathrm{E}=\mathrm{Bvl}$ $\begin{aligned} & R_e=\frac{6 \times 6}{6+6}+1=4 \Omega \\ & \therefore I=\frac{E}{R_e}=\frac{B v l}{R_e} \\ & \Rightarrow 10^{-3}=\frac{2 \times v \times 0.1}{4} \\ & \therefore \quad v=2 \mathrm{~cm} / \mathrm{s} \end{aligned}$

Hence, the correct answer is option: 'B'.

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