A) 9 mW
B) 18 Mw
C) 27 mW
D) 36 mW
Correct Answer: B
Solution :
[b]: The maximum value of the induced current, \[{{I}_{\max }}=\frac{\varepsilon \max }{R}=\frac{ABM\omega }{R}\]. Given, \[A=\pi {{r}^{2}}=3.14{{(8\times {{10}^{-2}}m)}^{2}},B=3\times {{10}^{-2}}T\], \[N=20,\omega =50\,\text{rad}\,{{s}^{-1}}\,\text{and}\,R=10\Omega \] \[\therefore \]\[{{I}_{\max }}=\frac{3.14{{\left( 8\times {{10}^{-2}}m \right)}^{2}}\left( 3\times {{10}^{-2}}T \right)\times 20\,(50\,\text{rad}\,{{\text{s}}^{\text{-l}}})}{10\Omega }\]\[=0.0603A\] The average power loss in the form of heat, \[{{P}_{av}}=\frac{1}{2}I_{\max }^{2}R=\frac{1}{2}{{(0.0603A)}^{2}}(10\Omega )\] \[=0.018\text{ }W=18\text{ }mW.\]
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