Question

A 11.3 m long steel beam is accidentally dropped by a construction crane from a height of 9.50 m. The horizontal component of the Earth's magnetic field over the region is 18.5 μT. What is the induced EMF in the beam just before impact with the Earth, assuming its long dimension remains in a horizontal plane, oriented perpendicularly to the horizontal component of the Earth's magnetic field?

Answer 1

Answer:

2.85 mV

Explanation:

L = 11.3 m

h = 9.5 m

B = 18.5 μ T = 18.5 x 10^-6 T

g = 9.8 m/s^2

u = 0

Let the velocity is v

v^2 = u^2 + 2 g h

v^2 = 2 x 9.8 x 9.5 = 186.2

v = 13.65 m/s

Motional emf, e = B v L = 18.5 x 10^-6 x 13.65 x 11.3 = 2.85 x 10^-3 V

e = 2.85 mV

Answer 2

The induced EMF in the beam just before impact with the earth will be E=2.85Mv

What will be the induced EMF in the beam just before impact with the Earth?

It is given that

Lenth=11.3m

Height = 9.50m

Earths magnetic field B=18.5$\mu T$

Now to calculate EMF

EMF =BvL

Here

B=Earths magnetic field =$18.5\times10^{-6}$

V= velocity of the falling object

L=Height from which the objet is falling

Now we first  calculate the velocity

$V^2=u^2+2gH$

$V^2=2\times9.8\times9.5$

$V=13.65 \ \dfrac{m}{s}$

Now from the EMF formula

$EMF=BvL$

$EMF=18.5\times10^{-6}\timnes13.65\times11.3$

$EMF=2.85Mv$

Thus the induced EMF in the beam just before impact with the earth will be E=2.85Mv

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