A) \[\frac{{{\mu }_{0}}I}{4\pi d}(\cos {{\phi }_{1}}+\cos {{\phi }_{2}})\]
B) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{2I}{d}\]
C) \[\frac{{{\mu }_{0}}I}{4\pi d}(\sin {{\phi }_{1}}+\sin {{\phi }_{2}})\]
D) \[\frac{{{\mu }_{0}}}{4\pi }\times \frac{I}{d}\]
Correct Answer: A
Solution :
Here, required angles\[{{\theta }_{1}}\]and\[{{\theta }_{2}}\]are\[(90{}^\circ -{{\phi }_{1}})\]an\[(90{}^\circ -{{\phi }_{2}})\] \[\therefore \]\[B=\frac{{{\mu }_{0}}I}{4\pi d}[\sin (90{}^\circ -{{\phi }_{1}})+\sin (90{}^\circ -{{\phi }_{2}})]\] \[=\frac{{{\mu }_{0}}I}{4\pi d}[\cos {{\phi }_{1}}+\cos {{\phi }_{2}}]\]
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