Verified Solution • 2025–26 Session moderate

QPages 249-253: Q29: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q29 Solution: Figure shows a part of an electric circuit. The wires AB, CD…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often calculate the force between only the adjacent wires (AB and CD) and forget that wire CD also experiences a repulsive force from wire EF. You must apply the principle of superposition by summing the forces from all other current-carrying wires to find the net force on a specific segment.

Figure shows a part of an electric circuit. The wires AB, CD and EF are long and have identical resistance. The separation between the neighbouring wires is 1.0 cm. The wires AE and BF have negligible resistance and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD.

Step-by-Step Verified Solution

**Answer:** F/LAB=1.2×10−4F/L_{AB} = 1.2 \times 10^{-4} N/m, F/LCD=0F/L_{CD} = 0. **Step-by-Step Explanation:** Ammeter reads 30 A. Currents in AB, CD, EF are 10 A, 10 A, 10 A (since identical resistances in parallel). Force per unit length between AB and CD: F=μ0×10×102π×0.01=2×10−4F = \frac{\mu_0 \times 10 \times 10}{2\pi \times 0.01} = 2 \times 10^{-4} N/m. Force on CD: from AB (attractive) and EF (repulsive) cancel. FCD=0F_{CD} = 0.

All Questions in this Chapter

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