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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______\mathrm{mV}. 

Option: 1

150


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{qVB}=\mathrm{qE}
\mathrm{E}=\mathrm{VB}
\Delta \mathrm{V}=\mathrm{EL}=\mathrm{VBL}
\Delta \mathrm{V}=2 \times 0.5 \times \frac{15}{100}=\frac{15}{100} \text { volt }
=15 \times 10^{-2} \mathrm{volt}
=150 \times 10^{-3} \mathrm{v}

Posted by

Pankaj Sanodiya

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