A circular loop of 20 turns carries a current of 5.0 A. If the magnetic field at the centre of the loop is 1.57 mT, the radius of the loop would be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4 cm
By the Biot-Savart law, the magnetic field at the centre of a circular coil of \( N \) turns and radius \( r \) is
\( \displaystyle B = \frac{\mu_0 N I}{2 r} \;\Rightarrow\; r = \frac{\mu_0 N I}{2 B} \)
\( \mu_0 = 4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}} \)
\( B = 1.57\,\mathrm{mT} = 1.57\times10^{-3}\,\mathrm{T} \)
\( \displaystyle r = \frac{4\pi\times10^{-7} \times 20 \times 5.0}{2 \times 1.57\times10^{-3}} \)
Since \( 2 \times 1.57 = 3.14 \approx \pi \),
\( \displaystyle r = \frac{4\pi\times10^{-7} \times 100}{\pi\times10^{-3}} = 4\times10^{-2}\,\mathrm{m} = 4\,\mathrm{cm} \)