JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 17)
The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance $${r \over 2}$$ from the centre is :
B/2
2B
$${\left( {{2 \over {\sqrt 5 }}} \right)^3}B$$
$${\left( {{2 \over {\sqrt 3}}} \right)^3}B$$
Explanation
$$B = {{{\mu _0}I} \over {2r}}$$
$${B_a} = {{{\mu _0}I{r^2}} \over {2\left( {{r^2} + {{{r^2}} \over 4}} \right)}}$$
$$ \Rightarrow {{{B_a}} \over B} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}$$
$$ \Rightarrow {B_a} = {\left( {{2 \over {\sqrt 5 }}} \right)^3}B$$
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