A) \[{{E}_{I}}>{{E}_{III}}={{E}_{V}}\,and\,\,\,{{E}_{II}}>{{E}_{IV}}\]
B) \[{{E}_{II}}={{E}_{IV}}={{E}_{V}}\,and\,\,\,{{E}_{I}}>{{E}_{III}}\]
C) \[{{E}_{I}}={{E}_{II}}={{E}_{III}}\,and\,\,\,{{E}_{IV}}>{{E}_{V}}\]
D) \[\frac{Mgx}{4}\]
Correct Answer: B
Solution :
At a distance x from the centre of the coil \[B=\frac{{{\mu }_{0}}In\,{{a}^{2}}}{2{{({{a}^{2}}+{{x}^{2}})}^{\frac{3}{2}}}}\] ; putting \[x=na\] \[B=\frac{{{\mu }_{0}}In\,{{a}^{2}}}{2{{a}^{3}}{{(1+{{n}^{2}})}^{\frac{3}{2}}}}=\frac{{{\mu }_{0}}{{n}_{0}}I}{2a{{(1+{{n}^{2}})}^{\frac{3}{2}}}}\] At the centre \[{{B}_{c}}=\frac{{{\mu }_{0}}{{n}_{0}}I}{2a}\] According to the questions \[{{2}^{-3/2}}\times \frac{{{\mu }_{0}}{{n}_{0}}I}{2a}=\frac{{{\mu }_{0}}{{n}_{0}}I}{2a{{(1+{{n}^{2}})}^{\frac{3}{2}}}}\] \[\Rightarrow \] \[{{(1+{{n}^{2}})}^{\frac{3}{2}}}={{2}^{\frac{3}{2}}}\] or \[n=1\]
You need to login to perform this action.
You will be redirected in
3 sec
