JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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.....\(mV\)
- A \(12\)
- B \(6\)
- C \(1\)
- D \(2\)
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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