cal magnetic force

2009-10-24 2:05 am

Two long parallel wires carry currents of 18 A and 5.4 A in opposite directions. The wires are separated by 0.15 m. What is the magnetic field midway (in teslas) between the two wires?
[Data: permeability of free space is 4xpix10-7T.m/A = 4piE-7 T.m/A.]
Calculate your answer to 2 significant figures with 1 decimal place and express it in the format used in data. ,

The answer = 6.2X10^5 T, Please show your solution steps by steps, thanks!!!

回答 (1)

天同

2009-10-24 4:17 am

✔ 最佳答案

Use the formaula: magnetic flux density B = u.I/(2.pi.R)
where u is the permeability of free space
I is the current in the wire
R is the distance from the wire
pi = 3.14159.....

Mangetic flux density due to 1the 8 A wire
= u(18)/[2.pi.(0.15/2)] T
Magnetic flux density due to 5.4 A wire
= u(5.4)/[2.pi.(0.15.2) T
Since the current in the two wires are flowing in opposite direction, the mangetic field lines at the mid-way between the wires are in the same direction. Hence,
Resultant magnetic flux density = u(18)/[2.pi.(0.15/2)] + u(5.4)/[2.pi.(0.15.2) T
= [4.pi.10^-7/2.pi.(0.15/2)].(18+5.4) T
= 6.2 x 10^-5 T

👍 0 👎 0