A long horizontal wire $P$ carries a current $50 A$ . It is rigidly fixed. Another line wire $Q$ is placed directly above and parallel to $P$ . The weight of wire $Q$ is $0.075 N m^{- 1}$ and caries a current of $25 A$ . Find the position of wire $Q$ from $P$ so that wire to remains suspended due to the magnetic repulsion-

3.33 m m

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33.3 m m

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334 m m

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333 m m

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Solution

The correct option is A $3.33 m m$
Formula,

$\frac{F}{l} = \frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2}}{r}$

$0.075 = 10^{- 7} \times \frac{2 \times 50 \times 25}{r}$

$\Rightarrow r = 3.33 \times 10^{- 3}$

$∴ r = 3.33 m m$

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The correct option is A 3.33 mmFormula,Fl=μ04π2I1I2r0.075=10−7×2×50×25r⇒r=3.33×10−3∴r=3.33mm

The correct option is A $3.33 m m$
Formula,

$\frac{F}{l} = \frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2}}{r}$

$0.075 = 10^{- 7} \times \frac{2 \times 50 \times 25}{r}$

$\Rightarrow r = 3.33 \times 10^{- 3}$

$∴ r = 3.33 m m$