Magnetic Effects & Magnetism

JEE Main level, single correct, medium

206 of 276

Question 206 of 276

A wire XX of length 50cm50cm carrying a current of 2A2A is placed parallel to a long wire YY of length 5m5m. The wire YY carries a current of 3A3A. The distance between two wires is 5cm5cm and currents flow in the same direction. The force acting on the wire YY is :
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Answer: (A) 1.2×10−5N1.2\times {10}^{-5}N directed towards wire XX.

An attractive force will act normal to XX towards YY because the direction of the currents in the wires is the same.
Force exerted on length ll due to the magnetic field is given as F=FXY=FYX=I1l1B12F={F}_{XY}={F}_{YX}={I}_{1}{l}_{1}{B}_{12}
⇒F=μ0I1I22πrl1\Rightarrow F=\dfrac{{\mu }_{0}{I}_{1}{I}_{2}}{2\pi r}{l}_{1}
Given here, I1=2A,I2=3A,l1=5m,r=5cm{I}_{1}=2A,{I}_{2}=3A,{l}_{1}=5m,r=5cm.
Putting the values, we have
F=4π×10−7×2×32π×5×10−2×50×10−2F=\dfrac{4\pi \times {10}^{-7}\times 2\times 3}{2\pi \times 5\times {10}^{-2}}\times 50\times {10}^{-2}
=1.2×10−5N=1.2\times {10}^{-5}N directed towards wire XX.

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