A conductor $ABOCD$ moves along its bisector with a velocity of $1\ m/s$ through a perpendicular magnetic field of $1\ wb/m^{2}$, as shown in fig. If all the four sides are of $1m$ length each, then the induced emf between points $A$ and $D$ is

Question

Dr. Abubacker · Accepted Answer

(b) There is no induced emf in the part $AB$ and $CD$ because they are moving along their length while emf induced between $B$ and $C\ i.e.,$ between $A$ and $D$ can be calculated as follows
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Induced emf between $B$ and $C =$ Induced emf between $A$ and $B = Bv\left( \sqrt{2}l ight) = 1 	imes 1 	imes 1 	imes \sqrt{2} = 1.41\ volt$

Q. A conductor $ABOCD$ moves along its bisector with a velocity of $1\ m/s$ through a perpendicular magnetic field of $1\ wb/m^{2}$, as shown in fig. If all the four sides are of $1m$ length each, then the induced emf between points $A$ and $D$ is

Explanation

Reviewed by Dr. Abubacker

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