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  2. Solve it, - Magnetic Effects of Current and Magnetism - JEE Main-3
# Engineering

Two wires A & B are carrying currents I_{1} & I_{2} as shown in the figure.

The separation between them is d.A third wire C carrying a current

I is to be kept parallel to them at a distance x from A such that the 

net force acting on it is zero. The possible values of x are:

  • Option 1)

    x=(\frac{I_{1}}{I_{1}-I_{2}})d   and     x=(\frac{I_{2}}{I_{1}+I_{2}})d

  • Option 2)

    x=(\frac{I_{2}}{I_{1}+I_{2}})d and  x=(\frac{I_{2}}{I_{1}-I_{2}})d

  • Option 3)

    x=(\frac{I_{1}}{I_{1}+I_{2}})d and   x=(\frac{I_{2}}{I_{1}-I_{2}})d

  • Option 4)

    x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d

 

Answers (1)
S solutionqc

 

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 

 

 

when I_2>I_1 ( that is wire 3 is on left side of both  1&2)

For equilibrium,

F_{31}=F_{32}

\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (d+x)}

\frac{d+x}{x}=\frac{I_2}{I_1}

=> x=(\frac{I_1}{I_2-I_1})d........................(1)

when I_1>I_2 ( that is wire 3 is on right side of both  1&2)

F_{31}=F_{32}

\frac{\mu _oI_1Il}{2\pi x}=\frac{\mu _oI_2Il}{2\pi (x-d)}

\frac{x-d}{x}=\frac{I_2}{I_1}

=>x=(\frac{-I_1}{I_2-I_1})d   .....................(2)

from (1) & (2)

x=\pm (\frac{I_1d}{I_1-I_2})


Option 1)

x=(\frac{I_{1}}{I_{1}-I_{2}})d   and     x=(\frac{I_{2}}{I_{1}+I_{2}})d

Option 2)

x=(\frac{I_{2}}{I_{1}+I_{2}})d and  x=(\frac{I_{2}}{I_{1}-I_{2}})d

Option 3)

x=(\frac{I_{1}}{I_{1}+I_{2}})d and   x=(\frac{I_{2}}{I_{1}-I_{2}})d

Option 4)

x=\pm (\frac{I_{1}}{I_{1}-I_{2}})d

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