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
A.
$$ - \frac{{2F}}{3}$$
B.
$$\frac{F}{3}$$
C.
$$ - 2F$$
D.
$$ - \frac{F}{3}$$
Answer :
$$ - \frac{{2F}}{3}$$
Solution :
Force between two long conductor carrying current,
$$\eqalign{
& F = \frac{{{\mu _0}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d} \times \ell \cr
& F' = - \frac{{{\mu _0}}}{{4\pi }}\frac{{2\left( {2{I_1}} \right){I_2}}}{{3d}}\ell \,\,\,\therefore \frac{{F'}}{F} = \frac{{ - 2}}{3} \cr} $$
Releted MCQ Question on Electrostatics and Magnetism >> Magnetic Effect of Current
Releted Question 1
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A.
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