CUET Preparation Today
Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:
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Zero $6×10^{-4}N$ toward left $9×10^{-4}N$ toward left $3×10^{-4}N$ toward right |
$3×10^{-4}N$ toward right |
Force due to first wire on middle wire is $F_1 = \frac{\mu_0 II_1}{6\pi}\times 25 = \frac{2500\mu_0}{2\pi}$ Force due to first wire on middle wire is $F_2 = \frac{\mu_0 II_2}{10\pi}\times 25 = \frac{1000\mu_0}{2\pi}$ Net Force F=F1−F2=$\frac{1500\mu_0}{2\pi}=3×10^{−4}N$ towards right |
