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 \[({{90}^{o}}-{{\phi }_{1}})\] \[({{90}^{o}}-{{\phi }_{2}})\] Hence, \[{{B}_{at}}0=\frac{{{\mu }_{0}}I}{4\pi d}[\sin ({{90}^{o}}-{{\phi }_{1}}]+\sin [{{90}^{o}}-{{\phi }_{2}}]\] \[\Rightarrow \] \[B=\frac{{{\mu }_{0}}I}{4\pi d}(\cos {{\phi }_{1}}+\cos {{\phi }_{2}})\]
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