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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_0I)/(2pid) (cos phi_1+cos phi_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 - A (a) Using `B=(mu_0i)/(4pia)[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_(n et)=(mu_0I)/(2pia)(cos phi_1+cos phi_2)` |
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