JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 22)
A 1 m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5$$\Omega$$, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be ____________ 10$$^{-3}$$ N.
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Explicação
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To move the rod with a constant velocity $v=4 \mathrm{~m} \mathrm{~s}^{-1}$
$\mathrm{F}_{\text {net }}$ on the rod should be zero.
$$ \begin{aligned} F & =B i l=0.15\left(\frac{B l v}{R}\right) l \\\\ & =0.15\left(\frac{0.15 \times 1 \times 4}{5}\right) \times 1 \\\\ & =0.03 \times 0.15 \times 4 \\\\ & =180 \times 10^{-4} \mathrm{~N} \\\\ & =18 \times 10^{-3} \mathrm{~N} \end{aligned} $$