JEE MAIN - Physics (2021 - 25th February Morning Shift - No. 5)
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ________.
1.0 m
0.15 m
0.2 m
0.1 m
Explanation
_25th_February_Morning_Shift_en_5_2.png)
$$B = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {x^2})}^{3/2}}}}$$
at x1 = 0.05 m, $${B_1} = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {{(0.05)}^2})}^{3/2}}}}$$
at x2 = 0.2 m, $${B_2} = {{{\mu _0}Ni{R^2}} \over {2{{({R^2} + {{(0.2)}^2})}^{3/2}}}}$$
$${{{B_1}} \over {{B_2}}} = {{{{({R^2} + 0.04)}^{3/2}}} \over {{{({R^2} + 0.0025)}^{3/2}}}}$$
$${\left( {{8 \over 1}} \right)^{2/3}} = {{{R^2} + 0.04} \over {{R^2} + 0.0025}}$$
4 (R2 + 0.0025) = R2 + 0.04
3R2 = 0.04 $$-$$ 0.0100
R2 = $${{0.03} \over 3}$$ = 0.01
R = $$\sqrt {0.01} $$ = 0.1 m
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