• # question_answer The wire shown in the figure carries a current of 40 A. If r = 3.14 cm, the magnetic field at point P will be A) $1.6\times {{10}^{-3}}T$              B) $3.2\times {{10}^{-3}}T$C) $4.8\times {{10}^{-3}}T$              D) $6.4\times {{10}^{-3}}T$

The straight portions of the wires do not contribute because the point P is along them. The field at P is due to 3/4th of the loop of radius r. Thus, $B=\frac{3}{4}\left( \frac{{{\mu }_{0}}I}{2r} \right)=\frac{3}{4}\times \frac{4\pi \times {{10}^{-7}}\times 40}{3.14\times {{10}^{-2}}}=4.8\times {{10}^{-3}}T$ Hence, the correction option is [c].