Question

A long horizontal wire contains a current such that 9.0✕1018 electrons per second flow through any given point from left to right. Determine the (a)current and (b)magnitude of the field of the long wire at a point 10 mm parallel to it in the north.

Answer 1

a.) $6.25 \times 10^{18}$ electrons per second gives current $= 1A$

Hence, $9 \times 10^{18}$ electrons gives current $= \dfrac{9 \times 10^{18}}{6.25 \times 10^{18}} = 1.44A$

b.) Magnetic field can be given as,

$B = \dfrac{\mu_0i}{2 \pi d}$

$B = \dfrac{1.25 \times 10^{-6} \times 1.44}{2 \times 3.14 \times 10 \times 10^{-3}}$

$B = 0.28 \times 10^{-2}T$