question_answer

A current loop, having two circular arcs joined by two radial lines y shown in the figure. It carries a current of 10 A. The magnetic field at point 0 will be close to: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) \[~1.5\,\,\times \,\,\text{1}{{0}^{-5}}\,T\]     

B) \[1.0\times {{10}^{-\,5}}\,T\]

C) \[1.5\,\,\times \,\,{{10}^{-7}}\,T~\]  

D) \[1.0\times \text{1}{{0}^{-7}}\,T\]

Correct Answer: B

Solution :

B due to wire QR: - \[{{\mu }_{0}}\,\frac{1}{8}\,\,\,\,\times \,\,\,\,\frac{10}{2\times (3\,cm)}\] B due to wire PS: - \[{{\mu }_{0}}\,\left( \frac{1}{8} \right)\,\,\,\,\times \,\,\,\,\frac{10}{2(5\,cm)}\] \[{{B}_{net}}\,=\,\,\frac{{{\mu }_{0}}}{8}\,\,\frac{10}{2}\,\,\left( \frac{1}{3\times {{10}^{-2}}}-\frac{1}{5\times {{10}^{-2}}} \right)\] \[=\,\,\,\frac{4\pi \,\,\times \,{{10}^{-7}}\,\times \,10\,}{16\,\,\times \,\,{{10}^{-2}}}\,\times \,\frac{2}{15}\,T\] \[=\,\,\,\frac{\pi \,\,\times \,\,{{10}^{-4}}}{30}\,\,T\] \[=\,\,\,\frac{3.14}{30}\,\,\times \,{{10}^{-\,4}}\,T\] \[=\,\,\,{{10}^{-\,5}}\,T\] Option is correct