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Magnetics question

2022 · 28 Jun · Shift 2 · Q62
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  5. /2022 · 28 Jun · Shift 2 · Q62

Magnetics question

2022 · 28 Jun · Shift 2 · Q62

JEE MainPhysicsMagneticsMCQ+4 / −1
Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 ×\times× 10 −-− 6 N, then the value of x is approximately :
  1. A
    1
  2. B
    2.4
  3. C
    1.4
  4. D
    2
View written solutionFree

Correct answer: C

  1. Use the force per unit length between two long parallel currents

For two long parallel wires separated by distance ddd, carrying currents I1I_1I1​ and I2I_2I2​:

FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}LF​=2πdμ0​I1​I2​​

Here,

  • I1=I2=xI_1 = I_2 = xI1​=I2​=x
  • d=0.20 md = 0.20\,\text{m}d=0.20m
  • FL=2×10−6 N/m\dfrac{F}{L} = 2 \times 10^{-6}\,\text{N/m}LF​=2×10−6N/m
  • In vacuum, μ0=4π×10−7 N/A2\mu_0 = 4\pi \times 10^{-7}\,\text{N/A}^2μ0​=4π×10−7N/A2

So,

2×10−6=(4π×10−7)x22π×0.202 \times 10^{-6} = \frac{(4\pi \times 10^{-7})x^2}{2\pi \times 0.20}2×10−6=2π×0.20(4π×10−7)x2​

  1. Simplify the expression

4π×10−72π=2×10−7\frac{4\pi \times 10^{-7}}{2\pi} = 2 \times 10^{-7}2π4π×10−7​=2×10−7

Thus,

2×10−6=2×10−7x20.202 \times 10^{-6} = \frac{2 \times 10^{-7} x^2}{0.20}2×10−6=0.202×10−7x2​

Since 0.20=150.20 = \frac{1}{5}0.20=51​,

2×10−70.20=2×10−7×5=10−6\frac{2 \times 10^{-7}}{0.20} = 2 \times 10^{-7} \times 5 = 10^{-6}0.202×10−7​=2×10−7×5=10−6

Therefore,

2×10−6=10−6x22 \times 10^{-6} = 10^{-6} x^22×10−6=10−6x2

  1. Solve for xxx

x2=2x^2 = 2x2=2

x=2≈1.414x = \sqrt{2} \approx 1.414x=2​≈1.414

So the current is approximately:

x≈1.4 Ax \approx 1.4\,\text{A}x≈1.4A

  1. Match with the options
  • A: 111
  • B: 2.42.42.4
  • C: 1.41.41.4
  • D: 222

Hence, the correct option is C.

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