A) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{10} \right)\]
B) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{8} \right)\]
C) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{6} \right)\]
D) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{4} \right)\]
Correct Answer: D
Solution :
Induced \[emf(e)\] \[=\frac{magnetic\,\,field\times change\,\,in\,\,area}{time}=\frac{B\Delta A}{t}\] Since, the circumference of the circular loop\[=2\pi r\], The side of the square loop\[=\frac{2\pi r}{4}=\frac{\pi r}{2}\] Therefore, \[\Delta A=\pi {{r}^{2}}-{{\left( \frac{\pi \,\,r}{2} \right)}^{2}}=\pi {{r}^{2}}\left( 1-\frac{\pi }{4} \right)\] \[\therefore \] \[e=\frac{B(\pi {{r}^{2}})}{t}\left( 1-\frac{\pi }{4} \right)\]You need to login to perform this action.
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