A) \[1.42\times {{10}^{-3}}A\]
B) \[2.67\times {{10}^{-3}}A\]
C) \[3.41\times {{10}^{-3}}A\]
D) \[4.21\times {{10}^{-3}}A\]
Correct Answer: B
Solution :
(b) \[2.67\times {{10}^{-3}}A\] Here, Area \[A={{l}^{2}}={{(12\,\,cm)}^{2}}=14\times {{10}^{-2}}{{m}^{2}}\] \[R=0.60\,\Omega ,\,{{B}_{1}}=0.10T,\,\theta =45{}^\circ ,\,{{B}_{2}}\] \[=0,\,dt=0.6\,s\] initial flux, \[{{\phi }_{1}}={{B}_{1}}A\,\cos \theta \] \[=0.10\times 14\times {{10}^{-2}}\times \cos \,45{}^\circ =9.9\times {{10}^{-4}}Wb\] Final flux, \[{{\phi }_{2}}=0\] Induced emf, \[\varepsilon =\frac{|d\phi |}{dt}=\frac{|{{\phi }_{2}}-{{\phi }_{1}}|}{dt}=\frac{|9.8\times {{10}^{-4}}|}{0.6\,s}\] \[=1.65\times {{10}^{-3}}V\] Current, \[l=\frac{\varepsilon }{R}=\frac{165\times {{10}^{-3}}}{0.6}=2.75\times {{10}^{-3}}A\]
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