Question #168424

A long, straight wire 9.27 mm in diameter carries 147 A distributed uniformly over its cross section.Find the magnitude of the magnetic field (also called the magnetic field strength)

(a) 2.50 mm from the wire’s axis and

(b) 7.50 mm from the axis


Expert's answer

a)


I=JA2πrB=μ0J(πr2)B=μ0Ir2R2B=4π107(147)2.51038(9.27103)2=6.72104 TI=JA\\2\pi rB=\mu_0 J(\pi r^2)\\B=\mu_0 I\frac{r}{2R^2} \\B=4\pi\cdot10^{-7} (147)\frac{2.5\cdot10^{-3}}{8(9.27\cdot10^{-3})^2}=6.72\cdot10^{-4}\ T

b)


I=JA2πrB=μ0J(πR2)B=μ0I1rB=4π107(147)17.5103=2.46102 TI=JA\\2\pi rB=\mu_0 J(\pi R^2)\\B=\mu_0 I\frac{1}{r} \\B=4\pi\cdot10^{-7} (147)\frac{1}{7.5\cdot10^{-3}}=2.46\cdot10^{-2}\ T


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