Question

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

Answer 1

Step-by-step explanation:

From Faraday's law,

`e m f &=-(d \phi)/(d t) \\ &=-(d)/(d t)(B A \cos \theta) \\ &=-(d)/(d t)(B l h \cos \theta) \\ &=-B l \cos \theta (d h)/(d t) \\ &=-B l v \cos 0^(\circ) \\ &=-B l v`

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

`e m f &=-B l(√(2 g h)) \\ 9.2 * 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 * 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 * 10^(-4) \mathrm{~T}`