A) \[\frac{{{\mu }_{0}}I}{2\pi d}(\cos {{\phi }_{1}}+\cos {{\phi }_{2}})\]
B) \[\frac{{{\mu }_{0}}I2I}{4\pi d}(tan{{\theta }_{1}}+\tan {{\theta }_{2}})\]
C) \[\frac{{{\mu }_{0}}I}{4\pi d}(sin{{\phi }_{1}}+\sin {{\phi }_{2}})\]
D) \[\frac{{{\mu }_{0}}I}{4\pi d}(cos{{\theta }_{1}}+\sin {{\theta }_{2}})\]
Correct Answer: A
Solution :
[a] Using \[B=\frac{{{\mu }_{0}}I}{4\pi d}\left[ \sin \,{{\theta }_{1}}+\sin \,{{\theta }_{2}} \right]\] But \[{{\theta }_{1}}+{{\phi }_{1}}={{90}^{o}}\] or \[{{\theta }_{1}}={{90}^{o}}-\phi \], \[\sin {{\theta }_{1}}=\sin ({{90}^{o}}-{{\phi }_{1}})=\cos {{\phi }_{1}}\] Similarly, \[\sin \,{{\theta }_{2}}=\cos \,{{\phi }_{2}}\] \[{{B}_{net}}=\frac{{{\mu }_{0}}I}{2\pi d}\left( \cos \,{{\phi }_{1}}+\cos \,{{\phi }_{2}} \right)\]
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