Answer to Question #168424 in Atomic and Nuclear Physics for Elizabethv

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


1
Expert's answer
2021-03-08T08:35:18-0500

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment