Question

tweleve wire of of equal length connected to form skeleton cube which moves with velocity v perpendicular to magnetic field B what will be the induced emf in each arm of the cube

Answer 1

For electromagnetic induction it is necessary that v, B and the length of the conductor sh be perpendicular to each other. If any two of them are parallel to each other then no current will be induced.

emf =
E=4Blv

Answer 2

Recollect that for generation of motional e.m.f., B, l and υ be along mutually perperdicular direction . If any or the two quantities are parallel,

e=0.

Velocity of all the 12 conductors is parallel to magnetic field B, therefore, no e.m.f. is induced in any conductor.

In fig. arms AD, BC, EH and FG are parallel to velocity (υ), So induced e.m.f. In these arms is zero. Further, arms AB, CD, EF, and GH are parallel to B→. Therefore, no e.m.f. is induced in these arms.

The rest of four arms AE, BF, CG and DH are perpendicular to both υ→ and B→. Therefore e.m.f. induced in each of these arms is

e=Blυ