Verified Solution • 2025–26 Session moderate

QPages 249-253: Q44: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q44 Solution: A circular coil of 200 turns has a radius of 10 cm and carries…

⚠️ Common Exam Traps & Conceptual Pitfalls

When setting $B_x = B_0/2$, the equation $(r^2+x^2)^{3/2}=2r^3$ must be raised to the $2/3$ power, giving $r^2+x^2=2^{2/3}r^2$, not $\sqrt[3]{2}\,r^2$. Then $x= r\sqrt{2^{2/3}-1}=r\sqrt{\sqrt[3]{4}-1}$. Forgetting the extra square‑root or using $\sqrt[3]{2}$ leads to an incorrect distance.

A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. a) Find the magnitude of the magnetic field B⃗\vec{B} at the centre of the coil. b) At what distance from the centre along the axis of the coil will the field BB drop to half its value at the centre? 43=1.5874...\sqrt[3]{4} = 1.5874...)

Step-by-Step Verified Solution

**Answer:** (a) B=1.26×10−3B = 1.26 \times 10^{-3} T (b) x=0.063x = 0.063 m. **Step-by-Step Explanation:** (a) N=200N = 200, r=0.1r = 0.1 m, i=2i = 2 A. B=μ0Ni2r=4π×10−7×200×22×0.1=2.51×10−3B = \frac{\mu_0 N i}{2r} = \frac{4\pi \times 10^{-7} \times 200 \times 2}{2 \times 0.1} = 2.51 \times 10^{-3} T. (b) Bx=μ0Nir22(r2+x2)3/2=B02B_x = \frac{\mu_0 N i r^2}{2(r^2 + x^2)^{3/2}} = \frac{B_0}{2}. (r2+x2)3/2=2r3⇒x=r(43−1)1/2=0.063(r^2 + x^2)^{3/2} = 2r^3 \Rightarrow x = r(\sqrt[3]{4} - 1)^{1/2} = 0.063 m.

All Questions in this Chapter

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