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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2019
  • Medium
  • Numerical

A solid metal cube of edge length \(2\, cm\) is moving in a positive \(y-\) direction at a constant speed of \(6\, m/s\). There is a uniform magnetic field of \(0.1\, T\) in the positive \(z-\) direction. The potential difference between the two faces of the cube perpendicular to the \(x-\) axis, is.....\(mV\)

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Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

\(E = vB\)
\(V = Ed = dvB\)\(=2\times 10^{-2}\times 6\times 0.1=12\;mV\)
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