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# Solve this problem - Magnetic Effects of Current and Magnetism - JEE Main-2

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

• Option 1)

$-2F$

• Option 2)

$F/3$

• Option 3)

$-2F/3$

• Option 4)

$-F/3$

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As we discussed in

Force between two parallel current carrying conductors -

$F=\frac{\mu }{4\pi } \frac{2I_{1 I_{2}}}{a} l$

$\frac{F}{l}=\frac{\mu o}{4\pi } \frac{2I_{1 I_{2}}}{a}$

- wherein

I1 and I2 current carrying two parallel wires

a-seperation between two wires

$:\; \; \; \; \; \; \; Initially, \; \; F= \frac{\mu _{0}}{2\pi }\frac{I_{1}I_{2}}{d}l$

$:\; \; \; \; \; \; \; Finally, \; \; F{}'= \frac{\mu _{0}}{2\pi }\frac{\left ( -2I_{1} \right )\left ( I_{2} \right )}{3d}l$

$\therefore \; \; \; \; \; \; \; Finally, \; \; \frac{F{}'}{F}= \frac{-\mu _{0}}{2\pi }\; \; \frac{ 2I_{1} I_{2} l}{3d}\times \frac{2\pi d}{\mu _{0}I_{1} I_{2}l}= -\frac{2}{3}$

$\therefore \; \; \; \; \; \; \; {F}'= -2F/3$

Option 1)

$-2F$

Incorrect option

Option 2)

$F/3$

Incorrect option

Option 3)

$-2F/3$

Correct option

Option 4)

$-F/3$

Incorrect option

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