Question

A long wire with a small current element of length 1 cm is placed at the origin and carries a current of 10 A along the X-axis. Find out the magnitude and direction of the magnetic field due to the element on  the Y-axis at a distance 0.5 m from it.

Answer 1

`vec|dB| = (mu_0)/(4 pi) (I dl sin theta ) / (r^2) `

dl = Δx = 10^-2 m , I = 10A r = 0.5 m = y, `mu_0 /4 pi = 10^(-7) (Tm)/A`

`theta = 90^circ ; sin theta = 1 `

`vec|dB| = (10^(-7) xx 10 xx 10^(-2))/(25 xx 10^(-2)) = 4 xx 10^(-8) T`

The direction of the field is in the + z-direction. This is so since,

`vec(dl) xx vec(r) = Delta x hat (i) xx y hat (j) = y  Delta x ( hat(i) xx hat(j) ) = y Delta x hat (k)` 

Hence the direction of magnetic field is towards positive z-axis.