Question:
If a wire of length 196 cm carrying a current of 0.98 A is bent in the form of a circular loop of two turns, then the magnetic field at the centre of the loop is approximately:
If a wire of length 196 cm carrying a current of 0.98 A is bent in the form of a circular loop of two turns, then the magnetic field at the centre of the loop is approximately:
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Magnetic field increases with the number of turns ($N$) and current ($I$).
Updated On: Jun 10, 2026
- 8 $\mu$T
- 4 $\mu$T
- 3 $\mu$T
- 6 $\mu$T
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The Correct Option is A
Solution and Explanation
Step 1: Concept
Magnetic field at the center of a circular coil: $B = \frac{\mu_0 N I}{2r}$.
Step 2: Analysis
Length $L = 196 cm = 1.96 m$. $N = 2$. $2\pi r N = L \implies r = L / (4\pi) = 1.96 / (4 \times 3.14) \approx 0.156 m$. $B = \frac{4\pi \times 10^{-7} \times 2 \times 0.98}{2 \times 0.156} \approx 8 \times 10^{-6} T = 8 \mu T$.
Step 3: Conclusion
The magnetic field is 8 $\mu$T.
Final Answer: (A)
Magnetic field at the center of a circular coil: $B = \frac{\mu_0 N I}{2r}$.
Step 2: Analysis
Length $L = 196 cm = 1.96 m$. $N = 2$. $2\pi r N = L \implies r = L / (4\pi) = 1.96 / (4 \times 3.14) \approx 0.156 m$. $B = \frac{4\pi \times 10^{-7} \times 2 \times 0.98}{2 \times 0.156} \approx 8 \times 10^{-6} T = 8 \mu T$.
Step 3: Conclusion
The magnetic field is 8 $\mu$T.
Final Answer: (A)
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