JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 13)
The magnetic field at the center of current carrying circular loop is $$B_{1}$$. The magnetic field at a distance of $$\sqrt{3}$$ times radius of the given circular loop from the center on its axis is $$B_{2}$$. The value of $$B_{1} / B_{2}$$ will be
9 : 4
12 : $$\sqrt5$$
8 : 1
5 : $$\sqrt3$$
Explanation
$${B_1} = {{{\mu _0}i} \over {2R}}$$
$${B_2} = {{{\mu _0}i{R^2}} \over {2{{({R^2} + {x^2})}^{{3 \over 2}}}}}$$
$$ \Rightarrow {{{B_1}} \over {{B_2}}} = {1 \over {{R^3}}}{({R^2} + {x^2})^{{3 \over 2}}}$$
$$ = {1 \over {{R^3}}}(8{R^3})$$
$$ = 8$$
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