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A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.Part1) Find the motional emf induced between the ends of the segment.
Part 2) What would the motional emf be if the wire segment was parallel to the y-axis?

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Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

[tex]E = (v\times B). L[/tex]

1)

As we know that

[tex]v = 18\hat i + 24\hat j + 72\hat k[/tex]

[tex]B = 0.080 \hat j[/tex]

[tex]L = 0.50 \hat k[/tex]

now we have

[tex]\vec v \times \vec B = 1.44\hat k - 5.76 \hat i[/tex]

so we have

[tex]E = 1.44 (0.50) = 0.72 V[/tex]

2)

If the length vector is along Y direction then we have

[tex]L = 0.50 \hat j[/tex]

so again we have

[tex]\vec v \times \vec B = 1.44\hat k - 5.76 \hat i[/tex]

so we have

EMF = 0

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