A) 3B
B) \[\sqrt{5}\]B
C) \[\frac{\sqrt{5}}{2}\]B
D) B
Correct Answer: B
Solution :
The point R will be in end-on position with respect to one dipole and in broad-side on position with respect to the other. So, \[{{B}_{P}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{M}_{1}}}{r_{1}^{3}}\] and \[{{B}_{Q}}=\frac{{{\mu }_{0}}}{4\pi }\frac{{{M}_{2}}}{r_{2}^{3}}\] Since, \[{{M}_{1}}={{M}_{2}}=M,\,\,{{r}_{1}}={{r}_{2}}=r\]and \[{{B}_{Q}}=B\] \[\therefore \] \[{{B}_{P}}=2B\] However, as \[{{B}_{P}}\]and \[{{B}_{Q}}\]are perpendicular to each other, \[B=\sqrt{B_{P}^{2}+B_{Q}^{2}}\] \[=\sqrt{{{(2B)}^{2}}+{{B}^{2}}}\] \[=\sqrt{5{{B}^{2}}}\] \[=\sqrt{5B}\]You need to login to perform this action.
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