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QJEEM19Ph9afc: JEE Main 2019 Physics - Class 12 JEEM19Ph9afc Solution: Two wires A & B are carrying currents I 1 & I 2 as shown in the …

Two wires A & B are carrying currents I1 & I2 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 : JEE Main 2019 (Online) 10th April Morning Slot Physics - Magnetic Effect of Current Question 150 English

Multiple Choice Options

  • A x=(I1I1+I2)d and x=(I2I1−I2)dx = \left( {{{{I_1}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d
  • B x=(I2I1+I2)d and x=(I2I1−I2)dx = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} - {I_2}}}} \right)d
  • C x=(I1I1−I2)d and x=(I2I1+I2)dx = \left( {{{{I_1}} \over {{I_1} - {I_2}}}} \right)d\,and\,x = \left( {{{{I_2}} \over {{I_1} + {I_2}}}} \right)d
  • D x=±I1dI1−I2x = \pm {{{I_1}d} \over {{I_1} - {I_2}}}

Step-by-Step Verified Solution

Net force on wire carrying current I per unit length is

μ0I1I2πx+μ0I2I2π(d−x)=0{{{\mu _0}{I_1}I} \over {2\pi x}} + {{{\mu _0}{I_2}I} \over {2\pi \left( {d - x} \right)}} = 0

I1x=I2x−d{{{I_1}} \over x} = {{{I_2}} \over {x - d}}

⇒x=I1dI1−I2\Rightarrow x = {{{I_1}d} \over {{I_1} - {I_2}}}

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