Question

An electron makes $3\times {10}^3$ revolutions per second in a circle of radius 0.5 angsfrom. Find the magnetic field B at the centre of the circle.

Answer 1

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

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

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

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

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