A conducting bar of 2m length is allowed to fall freely from a 50m high tower, keeping it aligned along the east-west direction. Find the emf induced in the rod when it is 20 m below the top of the tower. `g = 10 "ms"^(-2)`. Horizontal component of earth's magnetic field is `0.7 xx 10^(-4)` T and angle of dip =`60^@`.

Explanation

h=50 m, l=2 m , d=20 m , `g=10 ms^(-2)`
`B_h=0.7xx10^(-4)T, theta = 60^@ , N=100`, emf = ?
From eqn of free fall body
In `2ad =v^2-v_0^2, v_0=0`
`2gd=v^2 " " [because a=g]`
`2xx10xx20=v^2`
`400=v^2`
`therefore` v=20 m/s
Now, induced emf according to Faraday.s law
`therefore epsilon=B_h vl` [`because` neglecting negative sign]
`therefore epsilon = B vl cos theta [ because B_h =B cos theta]`
`therefore epsilon = 0.7xx10^(-4)xx20xx2xxcos 60^@ [ because cos 60^@ =1/2]`
=1.4 mV