A) \[\frac{1}{(2x-a)\,(2x+a)}\]
B) \[\frac{1}{{{x}^{2}}}\]
C) \[\frac{1}{{{(2x-a)}^{2}}}\]
D) \[\frac{1}{{{(2x+a)}^{2}}}\]
Correct Answer: A
Solution :
[a] \[{{\vec{B}}_{1}}=\frac{{{\mu }_{o}}I}{2\pi \left[ x-\frac{a}{2} \right]}=\frac{{{\mu }_{o}}I}{\pi \,[2x-a]}\] \[{{\vec{B}}_{2}}=\frac{{{\mu }_{o}}I}{2\pi \left[ x+\frac{a}{2} \right]}=\frac{{{\mu }_{o}}I}{\pi \,[2x+a]}\] \[emf=\frac{d\phi }{dt}=\frac{d(BA)}{dt}(\phi =BA)\] \[=\frac{AdB}{dt}\] \[={{a}^{2}}[{{\vec{B}}_{1}}-{{\vec{B}}_{2}}]\] \[={{a}^{2}}\,\left[ \frac{{{\mu }_{o}}I}{\pi \,[2x-a]}-\frac{{{\mu }_{o}}I}{\pi \,[2x+a]} \right]\] \[=a\frac{{{\mu }_{o}}I}{\pi }{{a}^{2}}\,\left[ \frac{11}{2x-a\,\,\,\,\,\,2x+a} \right]\] \[=a\frac{{{\mu }_{o}}I}{\pi }{{a}^{2}}\,\left[ \frac{2x+a-(2x-a)}{(2x-a)\,(2x+a)} \right]\] \[=a\frac{{{\mu }_{o}}I}{\pi }{{a}^{2}}\,\left[ \frac{2a}{(2x-a)\,(2x+a)} \right]\]
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