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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^@. |
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Answer» Solution :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 |
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