When a rod of length l is rotated with angular velocity of `omega` in a perpendicular field of induction B , about one end , the emf across its ends is

A rod of length l is rotated about its one end, perpendicular to a magnetic field of induction B What is the e.m.f induced in the rod

A conducting wire of length l is rotating with angular velocity w in a magnetic field B perpendicular to it. Prove that the induced e.m.f. at the two ends of the conducting wire is e = 1/2 Bwl^2 .

A conducting rod of length L rotates with angular speed omega in a uniform magnetic field of induction B which is perpendicular to its motion .The induceed emf developed between the two ends of the rod is

A rod of length l rotates with a uniform angular velocity omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is

A rod of length l rotates with a uniform angular velocity omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the lrod is

A rod of length l rotates with a uniform angular velocity omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the lrod is

A rod of length l rotates with a uniform angular velocity omega about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the lrod is