question_answer

Magnetic field at the centre of a coil in the form of a square of side 2 cm carrying a current of 1.414 A is

A)  \[8\times {{10}^{-5}}T\]                              

B)  \[8\times {{10}^{-5}}T\]

C)  \[1.5\times {{10}^{-5}}T\]                           

D)  \[6\times {{10}^{-5}}T\]

Correct Answer: A

Solution :

                 \[{{B}_{center}}=\frac{4\times {{\mu }_{0}}}{4\pi }\times \frac{I}{(a/2)}(\sin {{45}^{o}}+\sin {{45}^{o}})\]                 \[=4\times \frac{{{\mu }_{0}}}{4\pi }\times \frac{2I}{a}\times \frac{2}{\sqrt{2}}\]                 \[=\frac{{{\mu }_{0}}I\times 2\sqrt{2}}{\pi a}\]                 \[=\frac{4\pi \times {{10}^{-7}}\times 1.414\times 2\times \sqrt{2}}{\pi \times 2\times {{10}^{-2}}}\]                 \[=8\times {{10}^{-5}}T\]