Derive an expression for induced emf when a conductor of length l
Answers
Explanation:
diagram is missing
Let Upper end of rod is Q and lower end of rod is P
Figure shows a rod of length L moving in a magnetic field B with a constant velocity V. The length of the rod is perpendicular to the magnetic field and the velocity is perpendicular to both magnetic field and the rod.The free electrons of the wire also move with this velocity V together with the random velocity they have in the rod.The magnetic force due to the random velocity is zero on average.Thus the magnetic field exerts an average force
F⃗ b=qV⃗ ×B⃗ on each free electron where q = -1.6*10-19 C is the charge on the electron. This force is towards QP and hence the free electrons will move towards P . Negative charge is accumulated at P and positive charge is accumulated at Q.An electrostatic field E is developed within the wire from Q to P. The field exerted a force F⃗ e=qE⃗ on each free electron .The charge keeps on accumulation until a situation comes when Fb=Fe .
∣∣qV⃗ ×B⃗ ∣∣=∣∣qE⃗ ∣∣orvB=E
After this, there is no resultant force on the free electrons on the wire PQ . The potential difference between the ends Q and P is V = EL = VBL