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a b c d |
d |
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=$\frac{1500\mu_0}{2\pi}$=3×10−4N towards right |
