A conductor of length 2 m carrying current 2 A is held parallel to an infinitely long conductor carrying current of 12 A at a distance of 100 mm, the force j on a small conductor is:

8.6 x

10^{- 5} N

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6.6 x

10^{- 5} N

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7.6 x

10^{- 5} N

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9.6 x

10^{- 5} N

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Solution

The correct option is A 9.6 x $10^{- 5} N$
Here, $I_{1}$ = 2 A, $I_{2}$ = 12 A, r = 100 mm = 0.1 m, l = 2 m
Now total force on length l of small conductor

$F = \frac{μ_{0} 2 I_{1} I_{2} l}{4 π r} = \frac{10^{- 7} \times 2 \times 2 \times 12 \times 2}{0.1}$

$F = 9.60 \times 10^{- 5} N$

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A conductor of length 2 m carrying current 2 A is held parallel to an infinitely long conductor carrying current of 12 A at a distance of 100 mm, the force j on a small conductor is:

The correct option is A 9.6 x 10−5NHere,I1 = 2 A,I2 = 12 A, r = 100 mm = 0.1 m, l = 2 mNow total force on length l of small conductorF=μ02I1I2l4πr=10−7×2×2×12×20.1F=9.60×10−5N

The correct option is A 9.6 x $10^{- 5} N$
Here, $I_{1}$ = 2 A, $I_{2}$ = 12 A, r = 100 mm = 0.1 m, l = 2 m
Now total force on length l of small conductor

$F = \frac{μ_{0} 2 I_{1} I_{2} l}{4 π r} = \frac{10^{- 7} \times 2 \times 2 \times 12 \times 2}{0.1}$

$F = 9.60 \times 10^{- 5} N$