A) \[8\times {{10}^{-9}}\] tesla
B) \[4\times {{10}^{-19}}\] tesla
C) \[2\times {{10}^{-9}}\] tesla
D) zero
Correct Answer: A
Solution :
\[B={{B}_{1}}+{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\left( \frac{{{m}_{1}}}{r_{1}^{2}}+\frac{{{m}_{2}}}{r_{2}^{2}} \right)\] as \[{{m}_{1}}={{m}_{2}}\] and \[{{r}_{1}}={{r}_{2}}\] So, \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{m}_{1}}}{r_{1}^{2}}={{10}^{-7}}\frac{2\times {{10}^{-4}}}{{{(5\times {{10}^{-2}})}^{2}}}\] \[=8\times {{10}^{-9}}T\]
You need to login to perform this action.
You will be redirected in
3 sec
