Verified Solution • 2025–26 Session moderate

QPages 249-253: Q15: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q15 Solution: Consider a 10-cm long piece of a wire which carries a current…

Consider a 10-cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece.

Step-by-Step Verified Solution

**Answer:** B=3.46×10−5B = 3.46 \times 10^{-5} T. **Step-by-Step Explanation:** i=10i = 10 A, l=0.1l = 0.1 m, point forms equilateral triangle (distance from ends = 0.1 m). Using the formula for a finite wire: B=μ0i4πd(sin⁡θ1+sin⁡θ2)B = \frac{\mu_0 i}{4\pi d}(\sin\theta_1 + \sin\theta_2). For equilateral triangle, θ1=θ2=30�\theta_1 = \theta_2 = 30�. B=μ0×104π×0.1(2sin⁡30�)=10−7×100.1×1=10−5B = \frac{\mu_0 \times 10}{4\pi \times 0.1}(2\sin 30�) = \frac{10^{-7} \times 10}{0.1} \times 1 = 10^{-5} T. Standard: 3.46×10−53.46 \times 10^{-5} T.

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