A) \[\left( \frac{{{d}^{2}}+{{a}^{2}}}{2d} \right)qB\]
B) \[\frac{{{a}^{2}}}{2{{d}^{2}}}qB\]
C) \[\frac{4{{a}^{2}}}{(a+d)}qB\]
D) \[\frac{({{a}^{2}}-{{d}^{2}})}{2d}qB\]
Correct Answer: A
Solution :
[a] \[d-R\,(1-\cos \theta )\Rightarrow R\cos \theta =R-d\] \[R\sin \theta =a\Rightarrow {{R}^{2}}={{R}^{2}}-2Rd+{{d}^{2}}{{a}^{2}}\] \[\Rightarrow \,\,\,R=\frac{{{d}^{2}}+{{a}^{2}}}{2d}\,\,\,\,\Rightarrow \,\,\,p=\frac{{{d}^{2}}+{{a}^{2}}}{2d}qB\]You need to login to perform this action.
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