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

Question

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.

Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?

in progress 0
Trung Dũng 4 years 2021-07-12T23:34:08+00:00 1 Answers 162 views 0

Answers ( )

  1. quangkhai
    0
    2021-07-12T23:35:56+00:00
    Reply

    Answer:

    Explanation:

    From the information given:

    The motional emf can be computed by using the formula:

    E = L^{\to}*(V^\to*\beta^{\to})

    E = L^{\to}*((x+y+z)*\beta^{\to})

    E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)

    E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))

    E = 0.50*((18*0.800)

    E = 0.72 volts

    According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.

    As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .

    Then the motional emf will be:

    E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))

    E = 0 (zero)

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )