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The straight wire `AB` carries a current `I`. The ends of the wire subtend angles `theta_1` and `theta_2` at the point `P` as shown in figure. The magnetic field at the point `P` is A. `(mu0I)/(4pia)(sintheta_1-sintheta_2)`B. `(mu_0I)/(4pia)(sintheta_1+sintheta_2)`C. `(mu_0I)/(4pia)(costheta_1-costheta_2)`D. `(mu_0I)/(4pia)(costheta_1-costheta_2)` |
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Answer» Correct Answer - A NA |
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