A) \[5\times {{10}^{-7}}T\]
B) \[\sqrt{5}\times {{10}^{-7}}T\]
C) \[{{10}^{-7}}T\]
D) None of these
Correct Answer: B
Solution :
With respect to 1st magnet, P lies in end side-on position \[\therefore {{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{2M}{{{d}^{3}}} \right)\] (RHS) With respect to 2nd magnet. P lies in broad side on position. \[\therefore \ {{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{M}{{{d}^{3}}} \right)\] (Upward) \[{{B}_{1}}={{10}^{-7}}\times \frac{2\times 1}{1}=2\times {{10}^{-7}}T,\ {{B}_{2}}=\frac{{{B}_{1}}}{2}={{10}^{-7}}T\] As B1 and B2 are mutually perpendicular, hence the resultant magnetic field \[{{B}_{R}}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{{{(2\times {{10}^{-7}})}^{2}}+{{({{10}^{-7}})}^{2}}}\] \[=\sqrt{5}\times {{10}^{-7}}T\]You need to login to perform this action.
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