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The magnetic field at O due to current in the infinite wire forming a loop as a shown in Fig. A. `(mu_(0)I)/(4pid)(cosphi_(1)+cosphi_(2))`B. `(mu_(0))/(4pi)(2I)/d(tan theta_(1)+tan theta_(2))`C. `(mu_(0))/(4pi)(I)/d(sin phi_(1)+sin phi_(2))`D. `(mu_(0))/(4pi)(I)/d(cos theta_(1)+cos theta_(2))` |
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Answer» Correct Answer - B `B=(mu_(0)I)/(4pid) (sin theta_(1)+sin theta_(2))` But `theta_(1)+phi_(1)=90^(@)` or `theta_(1)=90^(@)-phi` `sin theta_(1)=sin(90^(@)-phi_(1))=cos phi_(1)` Similarly, `sin theta_(2)=cos phi_(2)` `B=(mu_(0)I)/(4pid) (cos phi_(1)+cos phi_(2))` |
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