A) equal
B) the same
C) doubled
D) quadrupled
Correct Answer: B
Solution :
[b] \[P=\frac{{{E}^{2}}}{R}=\frac{\pi {{r}^{2}}}{\rho \ell }{{\left( \frac{d\phi }{dt} \right)}^{2}}=\frac{\pi {{r}^{2}}}{\rho \ell }\left[ \frac{d}{dt}{{(NBA)}^{2}} \right]\] \[=\frac{\pi {{r}^{2}}}{\rho \ell }{{N}^{2}}{{A}^{2}}{{\left( \frac{dB}{dt} \right)}^{2}}\Rightarrow P\propto \frac{{{N}^{2}}{{r}^{2}}}{\ell }\] Case 1: \[{{P}_{1}}\propto \frac{{{N}^{2}}{{r}^{2}}}{\ell }\] Case 2: \[{{P}_{2}}\propto \frac{{{(4N)}^{2}}{{(r/2)}^{2}}}{4\ell }\] Note: When we decrease the radius of the wire, its length increases but volume remains the same \[\Rightarrow \,\,\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{1}\] \[\therefore \] Power remains the same.
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