A) 0.08 V
B) 0.16 V
C) 1.6 V
D) 0.04 V
Correct Answer: D
Solution :
From Faradays law of electromagnetic induction, the induced emf (e) is given by \[e=-\frac{\Delta \phi }{\Delta t}n\] ... (i) where \[\Delta \phi \] is rate of change of magnetic flux, given by \[\Delta \phi =B\,.\,\,A=BA\cos \theta \] ... (ii) where B is magnetic field intensity and A is area. Since coil is turned through 180° change in flux is \[\Delta \phi =BA-(-BA)=2BA\] \[\therefore \] \[e=-\frac{2\,BAn}{\Delta t}\] Given, \[B=0.4\times {{10}^{-4}}T\] \[A=500\,c{{m}^{2}}\] \[=500\times {{({{10}^{-2}})}^{2}}{{m}^{2}}\] \[\Delta t=\frac{1}{10}s\], n = 1000 \[e=-\frac{2\times 0.4\times 500\times {{10}^{-4}}\times 1000\times {{10}^{-4}}}{1/10}\] \[\Rightarrow \] \[e=0.04\]
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