Question #168415

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


1
Expert's answer
2021-03-04T17:24:34-0500

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


B=μ0I2R,B=\dfrac{\mu_0I}{2R},B=4π107 NA2650103 A21.0102 m=40.8106 T=40.8 μT.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=12μ0IR2(x2+R2),B=\dfrac{1}{2}\dfrac{\mu_0IR^2}{(x^2+R^2)},B=124π107 NA2650103 A(1.0102 m)2((20102 m)2+(1.0102 m)2)3/2,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=5.08109 T=5.08 nT.B=5.08\cdot10^{-9}\ T=5.08\ nT.

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