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A tong solenoid with 15 turns per em has a small loop of area 2.0 cm^(2) placed inside the solenoid normal to its axis, If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the inducced emf in the loop while the current is changing? |
| Answer» Solution :The magnetic field along the axis of solenoid is `B=mu_(0) n i` where n is no. of turns per unit length. flux through the smaller loop PLACED in solenoid is `phi=B xx A` SINCE current in solenoid is changing, emf induced in loop is `E=(d phi)/(dt)=(d)/(dt)[mu_(0)n i A]` `e=mu_(0)nA((di)/(dt))=4 pi xx 10^(-7) xx 15 xx 10^(2) xx 2 xx 10^(-4) xx ((4-2)/(1-0))=0.75 xx 10^(-6)V` | |