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Magnetics question

2022 · 26 Jun · Shift 2 · Q69
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  5. /2022 · 26 Jun · Shift 2 · Q69

Magnetics question

2022 · 26 Jun · Shift 2 · Q69

JEE MainPhysicsMagneticsNumerical+4 / −1
Two 10 cm long, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10 −-− 5 N, then separation between the wires is ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
View written solutionFree

Correct answer: 5

  1. Use the force between two parallel current-carrying wires

For two long parallel wires separated by distance ddd, the magnetic force on length LLL is

F=μ0I1I2L2πdF = \frac{\mu_0 I_1 I_2 L}{2\pi d}F=2πdμ0​I1​I2​L​

Here,

  • I1=I2=5 AI_1 = I_2 = 5\,\text{A}I1​=I2​=5A
  • L=10 cm=0.1 mL = 10\,\text{cm} = 0.1\,\text{m}L=10cm=0.1m
  • F=10−5 NF = 10^{-5}\,\text{N}F=10−5N
  • μ0=4π×10−7 N/A2\mu_0 = 4\pi \times 10^{-7}\,\text{N/A}^2μ0​=4π×10−7N/A2
  1. Substitute the values
10−5=(4π×10−7)(5)(5)(0.1)2πd10^{-5} = \frac{(4\pi \times 10^{-7})(5)(5)(0.1)}{2\pi d}10−5=2πd(4π×10−7)(5)(5)(0.1)​
  1. Simplify

First,

(5)(5)(0.1)=2.5(5)(5)(0.1) = 2.5(5)(5)(0.1)=2.5

So,

10−5=4π×10−7×2.52πd10^{-5} = \frac{4\pi \times 10^{-7} \times 2.5}{2\pi d}10−5=2πd4π×10−7×2.5​

Cancel π\piπ and simplify 42=2\frac{4}{2}=224​=2:

10−5=2×2.5×10−7d10^{-5} = \frac{2 \times 2.5 \times 10^{-7}}{d}10−5=d2×2.5×10−7​ 10−5=5×10−7d10^{-5} = \frac{5 \times 10^{-7}}{d}10−5=d5×10−7​
  1. Solve for ddd
d=5×10−710−5=5×10−2 md = \frac{5 \times 10^{-7}}{10^{-5}} = 5 \times 10^{-2}\,\text{m}d=10−55×10−7​=5×10−2m d=0.05 m=5 cmd = 0.05\,\text{m} = 5\,\text{cm}d=0.05m=5cm
  1. Final answer

The separation between the wires is

5\boxed{5}5​

cm.

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