Verified Solution • 2025–26 Session moderate

QPages 249-253: Q33: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Due to a Current Pages 249-253: Q33 Solution: A current-carrying circular coil of 100 turns and radius 5.0 cm …

A current-carrying circular coil of 100 turns and radius 5.0 cm produces a magnetic field of 6.0×10−5 T6.0 \times 10^{-5} \text{ T} at its centre. Find the value of the current.

Step-by-Step Verified Solution

**Answer:** i=0.48i = 0.48 A. **Step-by-Step Explanation:** N=100N = 100, r=0.05r = 0.05 m, B=6×10−5B = 6 \times 10^{-5} T. B=μ0Ni2r⇒i=2rBμ0N=2×0.05×6×10−54π×10−7×100=0.48B = \frac{\mu_0 N i}{2r} \Rightarrow i = \frac{2rB}{\mu_0 N} = \frac{2 \times 0.05 \times 6 \times 10^{-5}}{4\pi \times 10^{-7} \times 100} = 0.48 A.

All Questions in this Chapter

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