Homework 4 Begin Date: 2/21/2017 5:30:00 PM Due Date: 3/5/2017 11:59:00 PM End D
ID: 1540121 • Letter: H
Question
Homework 4 Begin Date: 2/21/2017 5:30:00 PM Due Date: 3/5/2017 11:59:00 PM End Date: 3/6/2017 4:00:00 AM (5%) Problem 15: An electric current is flowing through a long cylindrical conductor with radius a 0.45 m. The current density J 75 A/m is uniform in the cylinder. In this problem we consider an imaginary cylinder with radius r around the axis AB. Randomized Variables a 0.45 m 75 Am IB Otheexpertta com A 14% Part (a) When r is less than a, express the current inside the imaginary cylinder in terms of r and J Grade Summary Deductions 100% Potential a 7 8 9 HOME Submissions Attempts remaining: 12 5% per attempt) 1 2 3 detailed view Submit Hint give up! Hints: 5% deduction per hint. Hints remaining: 2 Feedback: 2% deduction per feedback. A 14% Part (b) Express B in terms of J and T A 14% Part (c) For r 0.5 a, calculate the numerical value of B in Tesla. A 14% Part (d) When r is greater than a, express the current inside the imaginary cylinder in terms of r, a and J A 14% Part (e) Express the magnitude of the magnetic field. B, at r a in terms of I and A 14% Part (f Express B in terms of J a and r A 14% Part (g For r -2 a, calculate the numerical value of B in TeslaExplanation / Answer
part a:
when r<a, current inside the maginary cylinder=J*area=J*pi*r^2
part b:
let magnetic field intensity be H.
as per ampere's law:
integration of H*dl =total current enclosed.
==>H*2*pi*r=J*pi*r^2
==>H=J*r/2
then B=mu*H=mu*J*r/2
where mu=magnetic permeability
part c:
given r=0.5*a=0.5*0.45=0.225 m
then B=mu*J*r/2
=4*pi*10^(-7)*7.5*0.225/2
=1.0603*10^(-6) T
part d:
when r >a , current =J*area=J*pi*a^2
part e:
using the same steps as part b,
H*2*pi*r=current enclosed=J*pi*a^2
==>H=J*a^2/(2*r)
then B=mu*H=mu*J*a^2/(2*r)
part f:
B= mu*J*r/2 ,for r<a
B=mu*J*a^2/(2*r), for r>a
part g:
for r=2*a=0.9 m
B=mu*J*a^2/(2*r)
=4*pi*10^(-7)*7.5*0.45^2/(2*0.9)=1.0603*10^(-6) T
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