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A 0.90-m steel bar is held with its length parallel to the east-west direction and dropped from a bridge. After it has fallen 32 m, the emf across its length is 9.2 x 10-4 V. Assuming the horizontal component of the earth's magnetic field points directly north, what is the magnitude of the horizontal component of the earth's magnetic field

Sagot :

Manetho

Explanation:

From Faraday's law,

[tex]e m f &=-\frac{d \phi}{d t} \\ &=-\frac{d}{d t}(B A \cos \theta) \\ &=-\frac{d}{d t}(B l h \cos \theta) \\ &=-B l \cos \theta \frac{d h}{d t} \\ &=-B l v \cos 0^{\circ} \\ &=-B l v[/tex]

Here, B is the magnetic field, l is the length of the rod, and v is the velocity of the rod.

 [tex]e m f &=-B l(\sqrt{2 g h}) \\ 9.2 \times 10^{-4} \mathrm{~V} &=-B(0.90 \mathrm{~m})\left(\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(32 \mathrm{~m})}\right) \\ B &=\frac{9.2 \times 10^{-4} \mathrm{~V}}{(0.90 \mathrm{~m})\left(\sqrt{2\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)(32 \mathrm{~m})}\right)} \\ &=0.408 \times 10^{-4} \mathrm{~T}[/tex]

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