Two long straight parallel wires carries currents of $200 mA$ and $1 A$ in opposite directions. If the wires are $20 cm$ apart, the distance of the neutral point from the $1 A$ wire is- (in $cm$ )

25

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35

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40

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55

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Solution

The correct option is A $25$

At, the neutral point,

$B_{1} = B_{2}$

$\frac{μ_{0} i_{1}}{2 π x}$ = $\frac{μ_{0} i_{2}}{2 π (r + x)}$

$\Rightarrow \frac{r + x}{x} = \frac{i_{2}}{i_{1}}$

$= \frac{1}{200 \times 10^{- 3}} = \frac{20 + x}{x}$

$\Rightarrow 5 = \frac{20 + x}{x}$

$\Rightarrow 20 + x = 5 x$

$\Rightarrow x = \frac{20}{4} = 5 cm$

Now, From $1 A$ conductor,

$d = r + x = 20 + 5 = 25 cm$

 Hence, $(A)$ is the correct answer.

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Two long straight parallel wires carries currents of 200 mA and 1 A in opposite directions. If the wires are 20 cm apart, the distance of the neutral point from the 1 A wire is- (in cm)

The correct option is A 25 At, the neutral point, B1=B2 μ0i12πx= μ0i22π(r+x) ⇒ r+xx=i2i1 =1200×10−3=20+xx ⇒ 5=20+xx ⇒ 20+x=5x ⇒ x=204=5 cm Now, From 1 A conductor, d=r+x=20+5=25 cm  Hence, (A) is the correct answer.

Two long straight parallel wires carries currents of $200 mA$ and $1 A$ in opposite directions. If the wires are $20 cm$ apart, the distance of the neutral point from the $1 A$ wire is- (in $cm$ )