A tightly wound 100 turn coil of radius 5 cm is carrying a certain amount of current. The magnitude of the magnetic field at the center of the coil is \( 2\pi\times10^{-4}\,\mathrm{T} \). The value of current flowing through the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.5\,\mathrm{A} \)
The magnetic field at the centre of a circular coil of \( N \) turns is \( B = \mu_0 N I/(2r) \).
\( \displaystyle I = \frac{2 r B}{\mu_0 N} \)
\( \displaystyle r = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( \displaystyle I = \frac{2 \times 0.05 \times 2\pi\times10^{-4}}{4\pi\times10^{-7} \times 100} = \frac{2\pi\times10^{-5}}{4\pi\times10^{-5}} = 0.5\,\mathrm{A} \)