An element `Delta l = Delta x hati ` is placed at the origin and carries a large current I = 10 A . What is the magnetic field on the y-axis at a distance of 0.5 m. `Deltax = 1 cm. `

Explanation

`|dB| = (mu_0)/(4pi) (Idl sin theta)/(r^2) `
`dl = Delta x = 10^(-2) m , I = 10 A , r = 0.5m = y , mu_0 // 4pi = 10^(-7) (TM)/(A)`
` theta = 90^@ , sin theta = 1`
`|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,
`dl xx r = Delta x hati xx y hatj = y Delta x (hati xx hatj) = y Delta x hatk`
We remind you of the following cyclic property of cross-products
`hati xx hatj = hatk , hatj xx hatk = hati , hatk xx hati = hatj`
Note that the field is small in magnitude.