Verified Solution • 2025–26 Session moderate

QPages 249-253: Q3: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q3 Solution: A copper wire of diameter 1.6 mm carries a current of 20 A.…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently substitute the given diameter ($1.6$ mm) directly into the formula $B = \frac{\mu_0 I}{2\pi r}$ without halving it to find the radius, resulting in an answer that is exactly half the correct value. Always verify that the linear dimension used in the denominator corresponds to the radial distance from the axis, not the full width of the conductor.

A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field B⃗\vec{B} due to this current.

Step-by-Step Verified Solution

**Answer:** Bmax=5×10−3B_{max} = 5 \times 10^{-3} T. **Step-by-Step Explanation:** i=20i = 20 A, r=0.8r = 0.8 mm =8×10−4= 8 \times 10^{-4} m. B=μ0i2πr=4π×10−7×202π×8×10−4=5×10−3B = \frac{\mu_0 i}{2\pi r} = \frac{4\pi \times 10^{-7} \times 20}{2\pi \times 8 \times 10^{-4}} = 5 \times 10^{-3} T.

All Questions in this Chapter

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