JEE MAIN - Physics (2004 - No. 29)
Two long conductors, separated by a distance $$d$$ carry current $${I_1}$$ and $${I_2}$$ in the same direction. They exert a force $$F$$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $$3d$$. The new value of the force between them is
$$ - {{2F} \over 3}$$
$${F \over 3}$$
$$-2F$$
$$ - {F \over 3}$$
Explanation
Force between two long conductor carrying current,
$$F = {{{\mu _0}} \over {4\pi }}{{2{I_1}{I_2}} \over d} \times \ell $$
$$F' = - {{{\mu _0}} \over {4\pi }}{{2\left( {2{I_1}} \right){I_2}} \over {3d}}\ell $$
$$\therefore$$ $${{F'} \over F} = {{ - 2} \over 3}$$
$$F = {{{\mu _0}} \over {4\pi }}{{2{I_1}{I_2}} \over d} \times \ell $$
$$F' = - {{{\mu _0}} \over {4\pi }}{{2\left( {2{I_1}} \right){I_2}} \over {3d}}\ell $$
$$\therefore$$ $${{F'} \over F} = {{ - 2} \over 3}$$
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