Question

The current carrying wire and the rod $AB$ are in the same plane. The rod moves parallel to the wire with a velocity $v$. Which one of the following statements is true about induced emf in the rod?

• A. End $A$ will be at lower potential w.r.t $B$.
• B. $A$ and $B$ will be at same potential.
• C. There will be no induced emf in the rod.
• D. Potential at $A$ will be higher than that at $B$.

Answer 1

The correct option is D Potential at $A$ will be higher than that at $B$.

Magnetic field is perpendicular to velocity into the plane containing rod and wire.
Total applied force (F) on negative charge $=-q(\overrightarrow{V}\times \overrightarrow{B})$
So,-ve charge moves along $B$ and +ve charge accumulates at $A$ . Therefore, end $A$ will be at high potential.