Answer to Question #168415 in Atomic and Nuclear Physics for bethv

Question

A single-turn wire loop is 2.0 cm in diameter and carries a 650 mA current.Find the magnetic field strength:

(a) at the loop centre

(b) on the loop axis, 20 cm from the centre

Answer 1

(a) Magnetic field strength at the loop centre can be found as follows:

$$B=\dfrac{\mu_0I}{2R},$$ $$B=\dfrac{4\pi\cdot10^{-7}\ \dfrac{N}{A^2}\cdot650\cdot10^{-3}\ A}{2\cdot1.0\cdot10^{-2}\ m}=40.8\cdot10^{-6}\ T=40.8\ \mu T.$$

(b) Magnetic field strength on the loop axis, 20 cm from the centre can be found as follows:

$$B=\dfrac{1}{2}\dfrac{\mu_0IR^2}{(x^2+R^2)},$$ $$B=\dfrac{1}{2}\cdot\dfrac{4\pi\cdot10^{-7}\ \dfrac{N}{A^2}\cdot650\cdot10^{-3}\ A\cdot(1.0\cdot10^{-2}\ m)^2}{((20\cdot10^{-2}\ m)^2+(1.0\cdot10^{-2}\ m)^2)^{3/2}},$$ $$B=5.08\cdot10^{-9}\ T=5.08\ nT.$$