A square conducting loop of side l, whose plane is perpendicular to the magnetic field, moves with velocity v perpendicular to the field. The two opposite sides of the loop that are perpendicular to its velocity lie in two mutually opposite uniform magnetic fields, each of strength B. The induced e…

Question

A square conducting loop of side l, whose plane is perpendicular to the magnetic field, moves with velocity v perpendicular to the field. The two opposite sides of the loop that are perpendicular to its velocity lie in two mutually opposite uniform magnetic fields, each of strength B. The induced emf in the loop is .

Accepted Answer

(2) 2Bvl. Given: a square loop of side l moving with velocity v; its two sides perpendicular to v are in opposite fields of magnitude B. Idea: a side of length l moving through a field has motional emf Bvl. The two sides parallel to v have no emf, because ⃗v × ⃗B is perpendicular to them. Leading side: emf Bvl, pushing charge one way along that side. Trailing side: the field is reversed, so its emf Bvl pushes charge the opposite way along its own side, which is the same sense of circulation round the loop. The two emfs are in series and aid each other: ε = Bvl + Bvl = 2Bvl. Check with flux: as the loop moves a distance dx, the area in one field grows by l dx and in the other shrinks by l dx, so the flux changes by 2Bl dx and ε = 2Blv. So the induced emf in the loop is 2Bvl.

Step-by-step solution

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