A) \[0{}^\circ \]
B) \[30{}^\circ \]
C) \[45{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: B
Solution :
In tan A positions \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2Md}{{{({{d}^{2}}-{{l}^{2}})}^{2}}}=H\tan \theta \] When \[d>>>l,\]then \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2{{M}_{d}}}{{{d}^{4}}}=H\tan {{\theta }_{A}}\] \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2M}{{{d}_{A}}^{3}}=H\tan {{\theta }_{A}}\] \[\left( \begin{align} & \text{Given:}\,{{\theta }_{A}}={{30}^{o}} \\ & {{d}_{A}}=18\,cm \\ \end{align} \right)\] \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times 2l\times {{m}_{A}}}{{{(18)}^{3}}}=H\tan {{30}^{o}}\] ....(i) In tan B position: \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{M}{{{({{d}^{2}}+{{l}^{2}})}^{3/2}}}=H\tan {{\theta }_{B}}\] \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{M}{{{d}^{3}}}=H\tan {{\theta }_{B}}\] \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2l\times {{m}_{B}}}{{{d}_{B}}^{3}}=\tan {{\theta }_{B}}H\] \[\left( \begin{align} & \text{Given:}{{\theta }_{B}}=? \\ & {{m}_{B}}=16\,{{m}_{A}} \\ & {{d}_{B}}=36\,cm \\ \end{align} \right)\] \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2l\times 16{{m}_{A}}}{{{(36)}^{3}}}=H\tan {{\theta }_{B}}\] Now, form Eqs. (i) and (ii),we have \[\frac{\tan {{30}^{o}}}{\tan {{\theta }_{B}}}=\frac{2}{{{(18)}^{3}}}\times \frac{{{(36)}^{3}}}{(16)}=1\] \[\tan {{\theta }_{B}}=\tan {{30}^{o}}\] \[{{\theta }_{B}}={{30}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec
