Respuesta :
Answer:
Induced emf in the rod is, [tex]\epsilon=0.08262\ V[/tex]
Explanation:
Given that,
Length of the rod, L = 22 cm = 0.22 m
Angle with x axis is 37.9 degrees with the positive x-axis and an angle of 52.1 degrees with the positive y-axis.
[tex]L=0.22(cos37.9\ i)+0.22(sin37.9\ j)[/tex]
[tex]L=(0.173\ i+0.135\ j)\ m[/tex]
Velocity of the rod, v = 6.8i m/s
Magnetic field, [tex]B=0.170i-0.24j-0.09k[/tex]
The formula for the emf induced in the rod is given by :
[tex]\epsilon=(v\times B){\cdot}L[/tex]
[tex]\epsilon=(6.8i\times (0.170i-0.24j-0.09k)){\cdot} (0.173\ i+0.135\ j)[/tex] Â Â Â
[tex]\epsilon=(0.612j-1.632k){\cdot}(0.173i+0.135j)[/tex] Â Â Â
[tex]\epsilon=0.08262\ V[/tex]
So, the magnitude of the emf induced in the rod is 0.08262 volts. Hence, this is the required solution. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
