Electro Magnetic Induction And Alternating Currents Question 524
Question: A solenoid has 2000 turns wound over a length of 0.30 metre. The area of its cross-section is $ 1.2\times {{10}^{-3}}m^{2} $ . Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is reversed in 0.25 sec, then the e.m.f. induced in the coil is [NCERT 1982; MP PMT 2003]
Options:
A) $ 6\times {{10}^{-4}}V $
B) $ 4.8\times {{10}^{-3}}V $
C) $ 6\times {{10}^{-2}}V $
D) 48 mV
Show Answer
Answer:
Correct Answer: D
Solution:
Induced emf $ e=M\frac{di}{dt}=\frac{{\mu_{0}}N_{1}N_{2}A}{l}.\frac{di}{dt} $
$ =\frac{4\pi \times {{10}^{-7}}\times 2000\times 300\times 1.2\times {{10}^{-3}}}{0.30}\times \frac{|2-(-2)|}{0.25} $ $ =48.2\times {{10}^{-3}}V=48\ mV $
