question_answer

Magnetic field at the centre of a coil in the form of a square of side \[2\text{ }cm\]carrying a current of \[1.414\text{ }A\]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 :

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