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InterviewSolution
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A long solenoid with 15 turns per cm has small loop of area `2.0 cm^(2)` placed inside, normal to the axis of the soleniod. If current carried by the solenoid changes steadily from 2 A to 4 A in 0.1 s, what is the induced voltage in the loop, while the current is changing ? |
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Answer» Number of turns on the solenoid = 15 turns/cm = 1500 turns/m Number of turns per unit length, n = 1500 turns The solenoid has a small loop of area, A = 2.0 `cm^(2)` = `2 × 10^(−4) m^(2)` Current carried by the solenoid changes from 2 A to 4 A. `therefore`Change in current in the solenoid, di = 4 − 2 = 2 A Change in time, dt = 0.1 s Induced emf in the solenoid is given by Faraday’s law as: `e=(dphi)/(dt)`....(i) Where, `phi`= Induced flux through the small loop = BA ... (ii) B = Magnetic field =`mu_(0)ni`.....(iii) `mu_(0)` = Permeability of free space = `4pi×10^(−7)` H/m Hence, equation (i) reduces to: `e=d/(dt)(BA)` `=Amu_(0)nxx((di)/(dt))` `=2xx10^(-4)xx4pixx10^(-7)xx1500xx2/0.1` `=7.54xx10^(-6)` V Hence, the induced voltage in the loop is `=7.54xx10^(-6)` |
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