Answer:
y = -0.42 m
1.16*10^-4 T(-k)
-1.73*10^4 N/C (j)
Explanation:
(a) Above the pair of wires, The field out of the page of the 50 A current will be stronger than the (—k) field of the 30 A current (k).
Between the wires, both produce fields into the page.
below the wires, y = - | y |
B = u_o*I/2πr (-k) + u_o*I/2πr (k)
0 = u_o/2πr[50/ | y |+0.28 (-k) + 30/| y | (k) ]
50 | y | = 30(| y | + 0.28)
| y | = -y
-50 y = 30*(0.28 - y)
y = -0.42 m
b) B = u_o*I/2πr (-k) + u_o*I/2πr (k)
B = 4π*10^-7/2π[ 50/0.28 -1 (-k) +30/1(-k) ]
= 1.16*10^-4 T(-k)
F = qv*B
F = (-2*10^-6)*(150*10^6(i) )(1.16*10^-4(-k))
F = 3.47*10^-2 N(-j)
c) F_e = qE
E = F_e/q
E = 3.47*10^-2/-2*10-6
= -1.73*10^4 N/C (j)
Answer:
Motion definition: the action or process of moving or being moved.
Examples:
1. moving a hand
2. riding a bicycle
3. running
Answer:
b) Demultiplexer.
Explanation:
The multiplexed signal occupies the total bandwidth of the communications channel, while that the constituent component signals, occupy only a small part of the bandwidth.
Prior to be demodulated, they must be separated each other, which is done (in a FDM multiplexing scheme) using a series of band-pass filters.
I believe the answer is true
