The horizontal component of the earth's magnetic field at a place is `3.0 X 10^(-4) T` and the dip is `53^@`. A metal rod of length 25cm is placed in the north - south direction and is moved at a constant speed of 10cm s^(-1) towards east. Calculate the emf induced in the rod.

Explanation

The inducedj emf is`epsilon= vBl` where l is the length of the rod, v is is its speed in the perpendicular direction and B is the component of the magnetic field perpendicular to both l and v
In the present case, B is the vertical component of the earth's magnetic field
the dip at a place is given by `tan delta = (B_v)/(B_H)` or `B_v = B_H tan delta` = (3
0 X 10^(-4) T) tan 53^@ = 4
0X10^(-4) T`
The emf induced is`epsilon = vBl` = (0
10ms^(-1))X (4
0X10^(-4) T) X (0
25m) =1
0 X 10^(-5) V = 10 mu V`