A) 8 F
B) 2 F
C) F / 2
D) F / 8
Correct Answer: D
Solution :
(d) F / 8. It is given that \[{{l}_{1}}={{l}_{1}}=l\] \[\therefore \] Force, \[F=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{l}^{2}}}{r}\] (1) Now, \[l'=l/2\] and \[r=2r\] So New force \[F;=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{(l/2)}^{2}}}{2r}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{l}^{2}}}{8r}\] \[=\frac{1}{8}\left( \frac{{{\mu }_{0}}}{4\pi }\frac{2{{l}^{2}}}{r} \right)=\frac{1}{8}F\] (from eq. (1))You need to login to perform this action.
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