question_answer

A wire loop PQRSP is constructed by joining two semi-circular coils of radii r1 and r2 respectively as shown in the figure. If the current flowing in the loop is 1, then the magnetic induction at the point O is                

A)                 \[\frac{{{\mu }_{0}}i}{4}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]

B)                 \[\frac{{{\mu }_{0}}i}{4}\left[ \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}} \right]\]

C)                 \[\frac{{{\mu }_{0}}i}{2}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]

D)                 \[\frac{{{\mu }_{0}}i}{2}\left[ \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}} \right]\]

Correct Answer: A

Solution :

                                                  In the following figure, magnetic field at O due to sections 1, 2, 3 and 4 are considered as \[{{B}_{1}},{{B}_{2}},{{B}_{3}}\]and \[{{B}_{4}}\] respectively                 \[{{B}_{1}}={{B}_{3}}=0\]                 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4}.\frac{i}{{{r}_{1}}}\]                 \[{{B}_{4}}=\frac{{{\mu }_{0}}}{4}.\frac{i}{{{r}_{2}}}\]                 So,          \[{{B}_{net}}={{B}_{2}}-{{B}_{4}}=\frac{{{\mu }_{i}}}{4}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{21}}} \right)\]