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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 Tesla

Explanation / 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