Magnetic Effects & Magnetism

JEE Main level, single correct, medium

185 of 276

Question 185 of 276

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire QQ is ____\_\_\_\_.

(μo=4π×10−7 T.m/A\mu_{\mathrm{o}}=4 \pi \times 10^{-7} \mathrm{~T}. \mathrm{m} / \mathrm{A})

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Answer: (D) 6×10−6 N6 \times 10^{-6} \mathrm{~N} towards RR

The force per unit length between parallel wires is F/L=μ0I1I22πdF/L = \frac{\mu_0 I_1 I_2}{2\pi d}.
Force on Q from P (opposite currents, repulsive towards R): FQPL=4π×10−7×3×12π×0.03=2×10−5\frac{F_{QP}}{L} = \frac{4\pi \times 10^{-7} \times 3 \times 1}{2\pi \times 0.03} = 2 \times 10^{-5} N/m.
Force on Q from R (same currents, attractive towards R): FQRL=4π×10−7×1×22π×0.02=2×10−5\frac{F_{QR}}{L} = \frac{4\pi \times 10^{-7} \times 1 \times 2}{2\pi \times 0.02} = 2 \times 10^{-5} N/m.
Total force per unit length towards R: 4×10−54 \times 10^{-5} N/m.
For 0.15 m length: F=4×10−5×0.15=6×10−6F = 4 \times 10^{-5} \times 0.15 = 6 \times 10^{-6} N towards R.

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