Question

A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and slide horizontally along the ground at a speed of 20.0 m/s. The velocity of the center of mass of the beam is northward while the length of the beam maintains an east–west orientation. The vertical component of the Earth's magnetic field at this location has a magnitude of 38.0 µT. What is the magnitude of the induced emf between the ends of the beam?

Answer 1

`emf= 9.88 * 10^(-3) T`

Step-by-step explanation:

Given:

• Length of the beam,
  `l=13\,m`

• speed of the beam,
  `v=20\,m.s^(-1)`

• magnitude of the vertical magnetic field,
  `B=38* 10^(-6) T`

According to the Faraday's law the emf induced in a rod passing transversely through a magnetic field is given as:

`emf= l.v.B`

`emf=13* 20* 38* 10^(-6)`

`emf= 9.88 * 10^(-3) T`