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NEET PHYSICSEasy

Two very long, straight, parallel conductors A and B carry current of 5 A and 10 A respectively and are at a distance of 10 cm from each other. The direction of the current in the two conductors is the same. The force acting per unit length between two conductors is: (μ0=4π×10−7\mu_0=4\pi \times 10^{-7} SI unit)

A

2×10−4 N m−12\times 10^{-4} \text{ N m}^{-1} and is attractive

B

2×10−4 N m−12\times 10^{-4} \text{ N m}^{-1} and is repulsive

C

1×10−4 N m−11\times 10^{-4} \text{ N m}^{-1} and is attractive

D

1×10−4 N m−11\times 10^{-4} \text{ N m}^{-1} and is repulsive

Step-by-Step Solution

  1. Identify Formula: The magnetic force per unit length (F/lF/l) between two infinite parallel current-carrying conductors is given by: Fl=μ0I1I22πd\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi d} where I1I_1 and I2I_2 are the currents and dd is the separation distance.
  2. Substitute Values: I1=5 AI_1 = 5 \text{ A} I2=10 AI_2 = 10 \text{ A} d=10 cm=0.1 md = 10 \text{ cm} = 0.1 \text{ m} μ0=4π×10−7 T m/A\mu_0 = 4\pi \times 10^{-7} \text{ T m/A} Fl=(4π×10−7)×5×102π×0.1\frac{F}{l} = \frac{(4\pi \times 10^{-7}) \times 5 \times 10}{2\pi \times 0.1}
  3. Calculate Magnitude: Fl=2×10−7×500.1=100×10−70.1=10−510−1=10−4 N m−1\frac{F}{l} = \frac{2 \times 10^{-7} \times 50}{0.1} = \frac{100 \times 10^{-7}}{0.1} = \frac{10^{-5}}{10^{-1}} = 10^{-4} \text{ N m}^{-1}
  4. Determine Direction:
  • Currents flowing in the same direction attract each other.
  • Currents flowing in opposite directions repel each other.
  • Since the problem states the direction is the same, the force is attractive.
  1. Conclusion: The force is 1×10−4 N m−11 \times 10^{-4} \text{ N m}^{-1} and is attractive.
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