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A metal rod `1/(sqrt(pi))` m long rotates about one of its ends in a plane perpendicular to a magnetic field of induction `4 xx 10^(-3)T`. If the emf induced between the ends of the rod is 16 mv, then the number of revolutions made by the rod per second isA. 3 rpsB. 4 rpsC. 5 rpsD. 6 rps |
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Answer» Correct Answer - B The induced emf `e = (d phi)/(dt) = d/(dt)(AB) = B(dA)/(dt)` In one rotation the rod traces out a circle of radius l or an area `A = pi l^(2)`. `:.` In n rotations it will trace out an area of `n pi l^(2)` `:. e = B(dA)/(dt) = B cdot (n pi l^(2))/(dt)` But `n/(dt) = ("No. of rev. performed")/("time") = "frequency" f` `:. e = B f pi l^(2) = Bf cdot A` `:. F = (e)/(B pi l^(2)) = (16 xx 10^(-3))/(4 xx 10^(-3) xx [pi((1)/(sqrt(pi)))^(2)])` `= 4/(pi xx 1/(pi)) = 4 r.p.s.` |
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