JEE MAIN - Physics (2023 - 11th April Evening Shift - No. 23)
A metallic cube of side $$15 \mathrm{~cm}$$ moving along $$y$$-axis at a uniform velocity of $$2 \mathrm{~ms}^{-1}$$. In a region of uniform magnetic field of magnitude $$0.5 \mathrm{~T}$$ directed along $$z$$-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be _________ mV.
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Answer
150
Explanation
$$
\begin{aligned}
& q E=e V B \\\\
& E=V B
\end{aligned}
$$
$$ \begin{aligned} \Delta V & =(E . d) \\\\ & =(V B) \times 0.15 \\\\ & =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\\\ & =0.15 \mathrm{~V}=150 \mathrm{mV} \end{aligned} $$
$$ \begin{aligned} \Delta V & =(E . d) \\\\ & =(V B) \times 0.15 \\\\ & =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\\\ & =0.15 \mathrm{~V}=150 \mathrm{mV} \end{aligned} $$
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