Electro Magnetic Induction And Alternating Currents Question 475
Question: A plane loop, shaped as two squares of sides a =1 m and b=0.4 m is introduced into a uniform magnetic field $ \bot $ to the plane of loop. The magnetic field varies as $ B={{10}^{-3}}\sin $ (100t) T. The amplitude of the current induced in the loop if its resistance per unit length is $ r=5,m{{\Omega }^{-1}} $ is
Options:
A) 2 A
B) 3 A
C) 4 A
D) 5 A
Show Answer
Answer:
Correct Answer: B
Solution:
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$ \phi $ (flux linked) $ =a^{2}B,\cos ,0^{o}-b^{2}B,\cos ,180^{o} $
$ E=-\frac{d\phi }{dt}=-(a^{2}-b^{2})\frac{dB}{dt} $
$ =(a^{2}-b^{2})B_{0},\omega \cos ,\omega t $ where $ B=B_{0} $ , $ \sin ,\omega t $ ,
$ B_{0}={{10}^{-3}}T $ ,
$ \omega =100 $
$ \therefore {I_{\max }}=(a^{2}-b^{2})\frac{B_{0}\omega }{R} $and $ R=(4a+4b)r=4(a+b)r $
$ \therefore {I_{\max }}=\frac{(a-b)B_{0}\omega }{4r}=\frac{(1-0.4)\times {{10}^{-3}}\times 100}{4\times 5\times {{10}^{-3}}}=3A $
