Physics genius!...heLp!?

Question

please help me solve these...i already answered #1 and #2.

http://s3.supload.com/free/Picture_252.jpg/view/

1. a small portion of parallel straight current-carrying conductors is shown in the figure. what is the flux density at point p due only to the 5-A current in the portion of wire 1 shown?

ans: 2.35 X 10^-6 Wb/m^2 (am i correct?)

2. in w/c direction does the flux at point p of #1 point?

ans: to the right of the page (am i correct?)

3. what is the flux density at point p in the figure due only to the 3-A current in the portion of wire 2 shown?

a. 1.64 X 10^-5 Wb/m^2

b. 2.45 X 10^-5 Wb/m^2

c. 3.11 X 10^-5 Wb/m^2

d. 3.65 X 10^-5 Wb/m^2

4. in w/c direction does the flux at point p of #3 point?

a. out of the page

b. to the left of the page

c. to the right of the page

d. into the page

5.what is the flux density at point p?

a. 0.10 X 10^-5 Wb/m^2

b. 0.55 X 10^-5 Wb/m^2

c. 0.71 X 10^-5 Wb/m^2

d. 1.36 X 10^-5 Wb/m^2

e. 3.39 X 10^-5 Wb/m^2

f. 3.99 X 10^-5 Wb/m^2

g. 4.80 X 10^-5 Wb/m^2

Answer 1

Ask your teacher why you are using Webers/m^2 instead of Teslas which is the modern equivalent.

1)By definition

1 Weber/m^2 = 1 Tesla

Use Ampere's Law to find B:

B = (u0) I /(2pi)r

The B field (magnetic flux density)due to the 5 amp current is

B = 3.33 x 10^-5 Tesla = 3.33 x 10^-5 Weber/m^2

Follow this method for the rest.

2)

By the right hand rule the B field points out of the page at point P.

3)

Flux density due to the 3 A current only is

B =2.99 x 10^-5 Webers/m^2

4) Again by the right hand rule the B field points Out of the Page.

5)

The total flux density is the sum of the two

B =  3.33 x 10^-5 + 2.99 x 10^-5 = 6.32 X 10^-5 Wb/m^2