Question

A 17.0 m long, thin, uniform steel beam slides south at a speed of 28.0 m/s. The length of the beam maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 30.0 µT. What is the magnitude of the induced emf between the ends of the beam (in mV)?

Answer 1

Answer:14.2 mV

Step-by-step explanation:

Given

Length of steel beam
`L=17\ m`

Magnetic field
`B=30\ \muT`

speed of beam
`v=28\ m/s`

Now consider as beam slides it encloses an area of rectangle of

width
`w=17\ m and length L'=v* t`

Area
`A=17v\cdot t`

and Induced EMF is
`V=-(d(BA))/(dt)`

`V=-B(d(17vt))/(dt)`

`V=-17Bv`

`V=-17* 30* 10^(-6)* 28`

`V=-14.2\ mV`

Magnitude of EMF
`=14.2\ mV`