JEE MAIN - Physics (2004 - No. 26)
A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $$B.$$ It is then bent into a circular loop of $$n$$ turns. The magnetic field at the center of the coil will be
$$2n$$ $$B$$
$${n^2}\,B$$
$$nB$$
$$2{n^2}\,B$$
Explanation
KEY CONCEPT : Magnetic field at the center of a circular coil of radius $$R$$ carrying
current is $$B = {{{\mu _0}i} \over {2R}}$$
Given: $$n \times \left( {2\pi r'} \right) = 2\pi R$$
$$ \Rightarrow nr' = R\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$B' = {{n.{\mu _0}i} \over {2r'}}\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
from $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$B' = {{n{\mu _0}i.n} \over {2\pi R}} = {n^2}B$$
current is $$B = {{{\mu _0}i} \over {2R}}$$
Given: $$n \times \left( {2\pi r'} \right) = 2\pi R$$
$$ \Rightarrow nr' = R\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
$$B' = {{n.{\mu _0}i} \over {2r'}}\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
from $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$B' = {{n{\mu _0}i.n} \over {2\pi R}} = {n^2}B$$
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