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How much charge will flow through a `200 Omega` galvanometer connected to a `400 Omega` circular coil of 1000 turns wound on a wooden stick 20 mm in diameter, if a magnetic field b = 0.0113 T parallel to the axis of the stick is decreased suddenly to zero. |
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Answer» Here, total resistance of the circuit, `R = 200 + 400 = 600 Omega` `n = 1000, D = 20 mm, r = 10 mm = 10^(-2) m`. `:.` Area of cross-section `A = pi r^(2) = pi (10^(-2))^(2) m^(2) = 10^(-4) pi m^(2)` `B_(1) = 0.0113 T, B_(2) = 0, :. Db = B_(2) - B_(1) = - 0.0113 T` Now `e = IR = (dq)/(dt) R` Also, `e = - (n d phi)/(dt) = (-n A dB)/(dt)` From (i) and (ii), `dq = (-n A dB)/(R ) = (- 1000 xx 10^(-4) pi xx (-0.0113))/(600) = + 5.9 xx 10^(-6) C` |
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