Respuesta :
Answer:
The induced emf between two end is [tex]34.02 \times 10^{-5}[/tex] V
Explanation:
Given:
Length of rod [tex]l = 1[/tex] m
Height [tex]h = 1.23[/tex] m
Magnetic field [tex]B = 6.93 \times 10^{-5}[/tex] T
For finding induced emf,
[tex]\epsilon = Blv[/tex]
Where [tex]v =[/tex] velocity of rod,
For finding the velocity of rod.
From kinematics equation,
[tex]v^{2} = v_{o} ^{2} + 2gh[/tex]
Where [tex]v_{o} =[/tex] initial velocity, [tex]g = 9.8 \frac{m}{s^{2} }[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2 \times 9.8 \times 1.23}[/tex]
[tex]v = 4.91[/tex] [tex]\frac{m}{s}[/tex]
Put the velocity in above equation,
[tex]\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91[/tex]
[tex]\epsilon = 34.02 \times 10^{-5}[/tex] V
Therefore, the induced emf between two end is [tex]34.02 \times 10^{-5}[/tex] V
