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JEE Main 2014 (11 Apr, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2014 (11 Apr, Shift 1)

Q56JEE Main 2014 · 11 APR · SHIFT 1Medium

Force on Current-Carrying Conductors

Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:

Figure

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Answer: (a)

  1. Fl=μ0I1I22πr\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}; opposite currents repel.
  2. 30 A wire (up) repels the 10 A wire (down) to the right: 2×10−7×30×100.03=2×10−3 N/m\frac{2\times10^{-7}\times30\times10}{0.03} = 2\times10^{-3}\text{ N/m}
  3. 20 A wire (up) repels it to the left: 2×10−7×20×100.05=8×10−4 N/m\frac{2\times10^{-7}\times20\times10}{0.05} = 8\times10^{-4}\text{ N/m}
  4. Net =1.2×10−3 N/m= 1.2\times10^{-3}\text{ N/m} to the right; F=1.2×10−3×0.25=3×10−4 NF = 1.2\times10^{-3}\times0.25 = 3\times10^{-4}\text{ N}
  5. So the answer is (a) 3×10−43\times10^{-4} N toward right.
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