Three long parallel straight conductors A, B and C carrying currents 3A, 1A and 2A respectively. The force experienced by the conductor ‘B’ of length 0.5 m is

10 \times 10^{- 6} N

from right to left

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10 \times 10^{- 6} N

from left to right

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5 \times 10^{- 6} N

from right to left

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5 \times 10^{- 6} N

from left to right

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Solution

The correct option is C $5 \times 10^{- 6} N$ from right to left
$F = \frac{μ_{0}}{2 π} [\frac{i_{1}}{r_{1}} - \frac{i_{2}}{r_{2}}] i L F = 2 \times 10^{- 7} [\frac{3}{3} - \frac{2}{4}] 1 \times 0.5 \times 10^{+ 2} = 2 \times 0.5 \times 10^{- 7} [1 - \frac{1}{2}] 10^{2} = 0.5 \times 10^{- 5}$
$= 5 \times 10^{- 6}$ from right to left.

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Three long parallel straight conductors A, B and C carrying currents 3A, 1A and 2A respectively. The force experienced by the conductor ‘B’ of length 0.5 m is

The correct option is C 5×10−6N from right to leftF=μ02π[i1r1−i2r2]iLF=2×10−7[33−24]1×0.5×10+2=2×0.5×10−7[1−12]102=0.5×10−5 =5×10−6 from right to left.