Answer:
e = 1.21 mV
Explanation:
given,
length of rod = 10 m
height of drop = 4.89 m
Earth’s magnetic field = 12.4 µT
acceleration of gravity = 9.8 m/s²
velocity of the beam
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2\times 9.8 \times 4.89}[/tex]
v = 9.79 m/s
emf of the beam
e = B l v
e = 12.4 x 10⁻⁶ x 9.79 x 10
e = 1.21 x 10⁻³ V
e = 1.21 mV