A) \[{{10}^{-3}}N\]
B) \[2.5\times {{10}^{-3}}N\]
C) zero
D) \[1.5\times {{10}^{-3}}N\]
Correct Answer: C
Solution :
Force on wire C due to wire D. \[{{F}_{1}}=\frac{{{\mu }_{0}}}{2\pi }\,\,\frac{{{I}_{1}}{{I}_{2}}}{r}l\] (repulsive) \[=2\times {{10}^{-7}}\times \frac{30\times 10}{3\times {{10}^{-2}}}\times 25\times {{10}^{-2}}\] \[=2\times {{10}^{-7}}\times 2500\] \[=5\times {{10}^{-4}}N\] Force on wire C due to wire G \[{{F}_{2}}=\frac{{{\mu }_{0}}}{2\pi }\,\frac{{{I}_{1}}{{I}_{2}}}{r}l\] (repulsive) \[=\frac{2\times {{10}^{-7}}\times 10\times 20}{2\times {{10}^{-2}}}\times 25\times {{10}^{-2}}\] \[=2\times {{10}^{-7}}\times 2500\] \[=5\times {{10}^{-4}}N\] Net force \[={{F}_{1}}-{{F}_{2}}\] \[=5\times {{10}^{-4}}N-5\times {{10}^{-4}}N\] \[=0\]
You need to login to perform this action.
You will be redirected in
3 sec
