Magnetic Flux through a Cylinder

[xxkylexx]

Homework Statement

A long, straight wire carrying a current of 4.00 A is placed along the axis of a cylinder of radius 0.500 m and a length of 3.00 m. Determine the total magnetic flux through the cylinder.

Homework Equations

Flux = BA
B(long wire) = (((4 * pi) * 10^-7) * I )/ (2 * pi * R)
A(cyclinder) = (2 * pi * r^2) + (2 * pi * r * h)

The Attempt at a Solution

I = 4
r = .5
h = 3

I tried to solve for B by saying:
B = (((4 * pi) * 10^-7) * 4 )/ (2 * pi * .5)
B = 1.6 * 10^-6

and then A:
A = (2 * pi * .5^2) + (2 * pi * .5 * 3)
A = 10.99Flux = (1.6 * 10^-6) * 10.99
Flux = 1.75 * 10^-5----------------------------This is incorrect. Not sure what I am doing wrong here. Any help is very much appreciated. This is due tonight at 11pm EST Thanks much,
Kyle

 

[sebaslu]

Notice here is asking you to find the total flux through the cylinder. However, the magnetic field lines are always perpendicular to the surface of the cylinder. So even if your calculations are right, it is not acting on the right direction. The final answer is zero

 

[paulfr]

The magnetic flux lines using the Right Hand Fist/Grip/Screw Rule ... circle around the wire perpendicular to the direction of the current. Since Flux is B dot A = B A cos theta, since theta is 90 degrees, the flux thru the cylinder is zero, 0. ...Theta is the angle between the normal to the surface and the flux lines of B = 90 degrees.

 

[kuruman]

The magnetic flux lines using the Right Hand Fist/Grip/Screw Rule ... circle around the wire perpendicular to the direction of the current. Since Flux is B dot A = B A cos theta, since theta is 90 degrees, the flux thru the cylinder is zero, 0. ...Theta is the angle between the normal to the surface and the flux lines of B = 90 degrees.

Please note that the OP's question is related to homework due at 11:00 pm EST on March 21, 2007.