Question

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

Answer 1

$B=\frac{{\mu }_{0}ni}{2R}$

$6.0\times {10}^{-7}=\frac{4\pi \times {10}^{-7}\times {10}^2\times i}{2\times 5\times {10}^{-2}}$

$i=\frac{60\times {10}^{-7}}{4\pi \times {10}^{-7}\times {10}^2}A$

$=4.777\times {10}^{-2}A$

$=47.77\times {10}^{-3}A\approx 48mA.$