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Answer :
Final answer:
The magnitude of the emf induced in the metal rod moving in a magnetic field is found to be approximately 0.359 V, using the factors of the magnetic field perpendicular to the rod's velocity, the velocity of the rod, and the length of the rod.
Explanation:
To calculate the magnitude of the electromotive force (emf) induced in a metal rod moving in a magnetic field, we can use the formula emf = Bvl sin(\theta), where B is the magnetic field strength, v is the speed of the rod, l is the length of the rod, and \theta is the angle between the magnetic field and the velocity vector. In this case, the rod is moving in the x-direction, so we need to find the component of the magnetic field that is perpendicular to the rod's velocity.
The velocity is in the positive x-direction, and we have the B-field components as B = (0.170T)i - (0.240T)j - (0.0900T)k. Since the rod makes angles with the x-axis and the y-axis, the effective component of B perpendicular to v will be a combination of the y and z components. However, because the rod is in the xy-plane, we can ignore the z component. Therefore, the perpendicular component of B is Bperp = -0.240T. The length of the rod is given as 22.0 cm or 0.220 m, and the speed is 6.80 m/s. Applying this, we get:
emf = Bperpvl
emf = 0.240T \* 6.80m/s \* 0.220m
emf = 0.35904 V
The magnitude of the emf induced in the rod is 0.35904 volts (or approximately 0.359 V when rounded to three significant figures).
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Rewritten by : Barada
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]
[tex]\epsilon=(0.612j-1.632k){\cdot}(0.173i+0.135j)[/tex]
[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.
