Derive the expression for the magnetic field due to a current-carrying coil of radius r at a distance x from the center along the X-axis.

#### Solution

Consider a conducting element dl of the loop. The magnitude dB of the magnetic field due to dl is given by the Biot-Savart law,

`"dB" = mu_°/(4pi) ("i"|vec(dl) xx vec("R")|)/"R"^3 `

`"dB" = mu_° /(4pi) ("idl")/"R"^2 (vec"idl" ⊥ vec"R")`

From the above figure, we can see, R^{2} = r^{2} +x^{2}

`"dB" = mu_° /(4pi) xx "idl"/("r"^2 + "x"^2)`

The direction of the magnetic field is shown in the figure, as we can see, only cosine component of the magnetic field will play a role here, all the sine component will get cancel out to give a zero net value.

`"dB"_"net" = mu_°/(4pi) xx "idl"/(("r"^2 + "x"^2)) xx costheta`

`"dB"_"net" = mu_°/(4pi) xx "idl"/(("r"^2 + "x"^2)) xx "r"/(sqrt("r"^2+"x"^2)) `

`"dB"_"net" = (mu_° "idl")/(4pi) "r"/(("r"^2 + "x"^2)^(3//2))`

∵ dl = rdθ

⇒ `"B"_"net" = (mu_°"i")/(4pi) ("r"^2 int_0^(2pi) "d"theta)/(("r"^2 + "x"^2)^(3//2)) = (mu_°"ir"^2)/(2("r"^2+ "x"^2)^(3//2))`

⇒ `"B"_"net" = (mu_°"ir"^2)/(2("r"^2+ "x"^2)^(3//2))`

⇒ `vec("B"_"net") = (mu_°"ir"^2)/(2("r"^2+ "x"^2)^(3//2)) hat"i"`