# Determine the magnitude and direction of the magnetic field at point a.

The figure below is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular application, the current in the inner conductor is I1 = 1.10 A out of the page and the current in the outer conductor is I2 = 3.04 A into the page. Assuming the distance d = 1.00 mm, answer the following.

(a) Determine the magnitude and direction of the magnetic field at point a.
 magnitude µT direction

(b) Determine the magnitude and direction of the magnetic field at point b.
 magnitude µT direction

asked Feb 19, 2012 in Physics

Applying Ampere's Law the magnitude of the magnetic field at a distance r from a long, straight wire carrying an electric current I is

Magnetic Field = [(permeability of free space)*(current)] / [2*pi*radius]

where permeability of free space= 4pi *10^-7  T*m/A

The magnitude of the magnetic field at point a is:

we are given:

Current 1 = 1.10 A

radius = 1 mm = 1*10^-3 m

plugging into the formula we get:

Magnetic Field at point a = [(4pi *10^-7  T*m/A)*(1.10 A)] / [2*pi*(1*10^-3 m)] =

Magnetic Field at point a = 2.2*10^-4 T  =  220  µT

The direction can be determined using the right hand rule, where your thumb points in the direction of the current, and the fingers point in the direction of the field.

After applyiing the right hand rule we find that the direction at point a is upward. (*current is out of the page is point thumb towards yourself, notice that your fingers are pointing upward at point a)

Similarily you can solve Part B, the only difference is that you must find the total current at point b, since there are two currents to consider now.

To find the total current simply add current 1 + current 2.

Notice that the two currents are in opposite directions, so what we do is subtract one from the other and say that the combined current is pointing in that direction.

So lets take current 2 to be positive.

Current 2 - Current 1 =  3.04 A - 1.10 A  = 1.94 A into the page (since the larger current, current 2, was into the page)

Now notice that the radius now has also changed.

it is now 3*r.

Now we can plug these values back into the formula and solve for the magnetic field at b.

Magnetic Field at point b = [(4pi *10^-7  T*m/A)*(1.94 A)] / [2*pi*(3*10^-3 m)] =

Magnetic Field at point b = 1.293*10^-4 T  = 129.3 µT

and the direction is downard (using right hand rule).

answered Feb 25, 2012 by ~Expert~ (1,318 points)
edited Feb 25, 2012 by Roar3215