JEE MAIN - Physics (2016 (Offline) - No. 11)
Two identical wires $$A$$ and $$B,$$ each of length $$'l'$$, carry the same current $$I$$. Wire $$A$$ is bent into a circle of radius $$R$$ and wire $$B$$ is bent to form a square of side $$'a'$$. If $${B_A}$$ and $${B_B}$$ are the values of magnetic fields at the centres of the circle and square respectively, then the ratio $${{{B_A}} \over {{B_B}}}$$ is:
$${{{\pi ^2}} \over {16}}$$
$${{{\pi ^2}} \over {8\sqrt 2 }}$$
$${{{\pi ^2}} \over {8}}$$
$${{{\pi ^2}} \over {16\sqrt 2 }}$$
Explanation
Case (a) :
_en_11_1.png)
$${B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi $$
$$ = {{{\mu _0}} \over {4\pi }}{I \over {\ell /2\pi }} \times 2\pi $$ $$\,\,\,\,\,\,\,\,\,$$ $$\left( {2\pi R = \ell } \right)$$
$$ = {{{\mu _0}} \over {4\pi }}{I \over \ell } \times {\left( {2\pi } \right)^2}$$
Case (b) :
_en_11_2.png)
$${B_B} = 4 \times {{{\mu _0}} \over {4\pi }}{I \over {a/2}}\,\,\,$$ $$\left[ {\sin \,\,{{45}^ \circ } + \sin \,\,{{45}^ \circ }} \right]$$
$$ = 4 \times {{{\mu _0}} \over {4\pi }} \times {I \over {\ell /8}} \times {2 \over {\sqrt 2 }}$$
$$ = {{{\mu _0}I} \over {4\pi \,\ell }} \times \root {32} \of 2 \,\,\,\,\,\,\left[ {4a = 1} \right]$$
$${B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi $$
$$ = {{{\mu _0}} \over {4\pi }}{I \over {\ell /2\pi }} \times 2\pi $$ $$\,\,\,\,\,\,\,\,\,$$ $$\left( {2\pi R = \ell } \right)$$
$$ = {{{\mu _0}} \over {4\pi }}{I \over \ell } \times {\left( {2\pi } \right)^2}$$
Case (b) :
$${B_B} = 4 \times {{{\mu _0}} \over {4\pi }}{I \over {a/2}}\,\,\,$$ $$\left[ {\sin \,\,{{45}^ \circ } + \sin \,\,{{45}^ \circ }} \right]$$
$$ = 4 \times {{{\mu _0}} \over {4\pi }} \times {I \over {\ell /8}} \times {2 \over {\sqrt 2 }}$$
$$ = {{{\mu _0}I} \over {4\pi \,\ell }} \times \root {32} \of 2 \,\,\,\,\,\,\left[ {4a = 1} \right]$$
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