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An air-cored solenoid with length 30 cm, area of cross-section 25 cm^(2) and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^(-3) s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid. |
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Answer» Solution :Inside a solenoid, the magnetic FIELD is given by `B = mu_(0)nI`, where n is the number of TURNS per unit length of the solenoid. The magnetic flux linked with the N turns of the solenoid of length I is `phi_(B) = NBA=N(mu_(0)nI)A=N(mu_(0)N/lI)A=(mu_(0)N^(2)IA)/l` `therefore` INDUCED emf `epsi=-(dphi_(B))/dt=-(mu_(0)N^(2)A)/l.(dI)/dt=(mu_(0)N^(2)A)/l((I_(1)-I_(2)))/t` As N = 500, l = 30 cm = 0.3m, A = 25 xx 10^(-4)m^(2), I_(1) = 2.5 A, I_(2) = 0 and t = 10^(-3)s. Here, `varepsilon = (4pi xx 10^(-7) xx (500)^(2)xx(25xx10^(-4)) xx (2.5-0))/((0.3)xx(10^(-3)))=6.5V. |
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