A) 2 V
B) 3V
C) 0.5 V
D) 6 V
Correct Answer: C
Solution :
(c) 0.5 V Here, area of coil \[A=0.1\text{ }m\times 0.05\text{ }m=5\times {{10}^{-3}}{{m}^{2}}\] Number of turns, \[N=100\] Initial flux linked with the coil \[{{\phi }_{1}}=BA\,\cos \,\theta =0.1\times 5\times {{10}^{-3}}\cos \,0{}^\circ =5\times {{10}^{-4}}Wb\] Final flux linked with the coil \[{{\phi }_{2}}=0.05\times 5\times {{10}^{-3}}\cos 0{}^\circ =25\times {{10}^{-5}}Wb\] \[=2.5\times {{10}^{-4}}Wb\] The magnitude of induced emf in the coil is \[\varepsilon =\frac{N|\Delta \,\phi |}{\Delta t}=\frac{N|{{\phi }_{2}}-{{\phi }_{1}}|}{t}\] \[=\frac{100|2.5\times {{10}^{-4}}-5\times {{10}^{-4}}|}{0.05}\] \[=\frac{100\times 2.5\times {{10}^{-4}}}{0.05}V=0.5\,V\]
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