JEE MAIN - Physics (2002 - No. 9)
If in a circular coil $$A$$ of radius $$R,$$ current $$I$$ is flowing and in another coil $$B$$ of radius $$2R$$ a current $$2I$$ is flowing, then the ratio of the magnetic fields $${B_A}$$ and $${B_B}$$, produced by them will be
$$1$$
$$2$$
$$1/2$$
$$4$$
Explanation
KEY CONCEPT : We know that the magnetic field produced by a current carrying circular coil of radius $$r$$
at its center is $$B = {{{\mu _0}} \over {4\pi }}{I \over r} \times 2\pi $$
Here $${B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi $$ and
$${B_B} = {{{\mu _0}} \over {4\pi }}{{2I} \over {2R}} \times 2\pi $$
$$ \Rightarrow {{{B_A}} \over {B{}_B}} = 1$$
at its center is $$B = {{{\mu _0}} \over {4\pi }}{I \over r} \times 2\pi $$
Here $${B_A} = {{{\mu _0}} \over {4\pi }}{I \over R} \times 2\pi $$ and
$${B_B} = {{{\mu _0}} \over {4\pi }}{{2I} \over {2R}} \times 2\pi $$
$$ \Rightarrow {{{B_A}} \over {B{}_B}} = 1$$
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