A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s^–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10^–4 Wb m^–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?

Answer –                                                             

Wire’s Length is given by l = 10 m

Speed of the falling of wire, v = 5.0 m/s

Strength of magnetic field is given by –

\[B=0.3\times {{10}^{-4}}Wb{{m}^{-2}}\]

(a) EMF induced in the wire,

e = Blv

$$$e=0.3\times {{10}^{-4}}\times 5\times 10=1.5\times {{10}^{-3}}V$

(b) Using Fleming’s right-hand thumb rule, we may calculate the direction of the induced current; in this case, the current is flowing from West to East.

(c) In this scenario, the wire’s eastern end will have a higher potential.