$A$ 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,$0.30 \times 10^{-4} \; Wb \, m^{-2}$.
$(a)$ 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?

(A) Given:
Length of the wire,$l = 10 \; m$
Falling speed of the wire,$v = 5.0 \; m \, s^{-1}$
Magnetic field strength,$B = 0.30 \times 10^{-4} \; Wb \, m^{-2}$
$(a)$ The instantaneous value of the induced $emf$ is given by the formula $\varepsilon = B \cdot l \cdot v$.
Substituting the values:
$\varepsilon = (0.30 \times 10^{-4} \; Wb \, m^{-2}) \times (10 \; m) \times (5.0 \; m \, s^{-1})$
$\varepsilon = 1.5 \times 10^{-3} \; V$
$(b)$ According to Fleming's Right-Hand Rule,the direction of the induced current (and thus the $emf$) is from West to East.
$(c)$ Since the induced current flows from West to East inside the wire,the eastern end of the wire is at a higher electrical potential.