A conducting slider PQ (length l) moves right at speed v on a stationary U-shaped conductor of width l in uniform perpendicular magnetic field B. Wire resistance per unit length is r. The induced circuit current is 4Blv/[Nr(l+vt)]. Find N.
Physics · Mixed Chapter
JEE Mains

Question

A conducting slider PQ (length l) moves right at speed v on a stationary U-shaped conductor of width l in uniform perpendicular magnetic field B. Wire resistance per unit length is r. The induced circuit current is 4Blv/[Nr(l+vt)]. Find N.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Motional emf: The sliding rod PQ has length ℓ, moves at speed v and cuts a perpendicular magnetic field B. Induced emf magnitude = Bℓv. Step 2 — Find total conducting length: At time t, horizontal extent of the rectangular loop = vt; its vertical extent = ℓ. Total wire length around the loop = 2(ℓ + vt). Step 3 — Circuit resistance: With resistance per unit length r, R = 2r(ℓ + vt). Step 4 — Induced current: I = emf/R = Bℓv/[2r(ℓ + vt)] = 4Bℓv/[8r(ℓ + vt)]. Step 5 — Match the supplied…