A) \[\frac{{{\mu }_{0}}{{i}_{1}}{{i}_{2}}}{4\pi \,{{r}^{2}}}\]
B) \[\frac{{{\mu }_{0}}{{i}_{1}}}{4\pi \,{{r}^{2}}}\]
C) \[\frac{{{\mu }_{0}}{{i}_{2}}}{4\pi \,{{r}^{2}}}\]
D) zero
Correct Answer: D
Solution :
Let \[{{i}_{1}}\] and \[{{i}_{2}}\] be the currents in the two parts and \[{{R}_{1}}\] and \[{{R}_{2}}\] be the resistances. As two parts are in parallel with each other, then \[{{i}_{1}}{{R}_{1}}={{i}_{2}}{{R}_{2}}\] Also \[{{R}_{1}}\propto {{l}_{1}}\] and \[{{R}_{2}}\propto {{l}_{2}}\] \[\therefore \] \[{{i}_{1}}{{l}_{1}}={{i}_{2}}{{l}_{2}}\] Now \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\,\frac{{{i}_{2}}{{i}_{2}}\sin \,{{90}^{o}}}{{{r}_{1}}^{2}}\] and \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\,\frac{{{i}_{2}}{{i}_{2}}\,\sin \,{{90}^{o}}}{{{r}_{2}}^{2}}\] As \[{{i}_{1}}{{l}_{1}}={{i}_{2}}{{l}_{2}}\] \[\therefore \] \[{{B}_{1}}={{B}_{2}}\] As directions of \[{{B}_{1}}\] and \[{{B}_{2}}\] are opposite, thus \[{{B}_{1}}-{{B}_{2}}=0\]
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