HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q10 Solution: Two parallel wires carry equal currents of 10 A along the same…

QPages 249-253: Q10: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q10 Solution: Two parallel wires carry equal currents of 10 A along the same…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently assume the angle between the magnetic field vectors is $90^\circ$ or $120^\circ$ by mistake; however, for an equilateral triangle configuration with currents in the same direction, the angle between the field vectors at the third vertex is exactly $60^\circ$, leading to a resultant magnitude of $B\sqrt{3}$ rather than $B$ or $B\sqrt{2}$.

Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.

Step-by-Step Verified Solution

**Answer:**

B = 8.66 \times 10^{-5}

T. **Step-by-Step Explanation:** Two wires 2 cm apart, each 10 A, same direction. Point P is 2 cm from each (equilateral triangle).

B_1 = B_2 = \frac{\mu_0 \times 10}{2\pi \times 0.02} = 10^{-4}

T. Angle between fields = 60�.

B_{net} = \sqrt{B_1^2 + B_2^2 + 2B_1B_2\cos 60�} = B\sqrt{3} = 1.73 \times 10^{-4}

QPages 249-253: Q10: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q10 Solution: Two parallel wires carry equal currents of 10 A along the same…

Step-by-Step Verified Solution

All Questions in this Chapter

Related Learning Resources & Exam Preparation