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A 0.5 m long solenoid of 10 turns/cm has area of cross-section 1 cm^(2). Calculate the voltage induced across its ends if the current in the solenoid is changed from 1 A to 2 A in 0.1 s. |
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Answer» Solution :Here length of SOLENOID I= 0.5 m, number of turns per unit length n = 10 turns/em = `10 xx 100 m^(-1) =1000 m^(-1),` area of cross-section `A = 1 cm^(2) = 10^(-4) m^(2)`, change in current `deltal = I_(2) - I_(1) = (2 - 1) A = 1 A` and time `DELTAT = 0.1 s.` `therefore` Self-inductance of solenoid `L = mu_(0)n^(2)lA = (4pi xx 10^(-7)) xx (1000)^(2) xx 0.5 (10^(-4) = 6.28 xx 10^(-5) H` `therefore` Magnitude of induced voltage, `|varepsilon|= L(DeltaI)/(Deltat) = 6.28 xx 10^(5) xx 1/(0.1)= 6.28 xx 10^(-4) V or 0.628 mV` |
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