A) \[40\times {{10}^{-7}}N\]
B) 8\[\pi \] \[\times {{10}^{-7}}N\]
C) \[8\times {{10}^{-5}}N\]
D) \[4\times {{10}^{-5}}N\]
Correct Answer: C
Solution :
Here, \[{{i}_{1}}=10A,\,\,\,{{i}_{2}}=2A\] \[r=10cm=0.1m,\,\,l=2m\] Using the relation \[F=\frac{{{\mu }_{0}}{{i}_{1}}{{i}_{2}}l}{2\pi R}\] or \[F=\frac{4\pi \times {{10}^{-7}}\times 10\times 2\times 2}{2\pi \times 0.10}\] So, \[F=8\times {{10}^{-5}}N\]You need to login to perform this action.
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