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)?

Respuesta :

Answer:14.2 mV

Explanation:

Given

Length of steel beam [tex]L=17\ m[/tex]

Magnetic field [tex]B=30\ \muT[/tex]

speed of beam [tex]v=28\ m/s[/tex]

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

width [tex]w=17\ m and length L'=v\times t [/tex]

Area [tex]A=17v\cdot t[/tex]

and Induced EMF is [tex]V=-\frac{d(BA)}{dt}[/tex]

[tex]V=-B\frac{d(17vt)}{dt}[/tex]

[tex]V=-17Bv[/tex]

[tex]V=-17\times 30\times 10^{-6}\times 28[/tex]

[tex]V=-14.2\ mV[/tex]

Magnitude of EMF[tex]=14.2\ mV[/tex]